Permutation and Combination Questions in English

(A) The total number of questions is $n = 6$.
The number of correct answers required is $r = 4$.
For each question,there is $1$ correct answer and $3$ incorrect answers.
The number of ways to choose $4$ questions out of $6$ to be correct is given by the combination formula ${}^{6}C_{4}$.
For the $4$ correct questions,there is only $1$ way to answer each correctly $(1^4)$.
For the remaining $6 - 4 = 2$ questions,each must be answered incorrectly,and there are $3$ choices for each incorrect answer $(3^2)$.
Therefore,the total number of ways is ${}^{6}C_{4} \times 1^4 \times 3^2 = 15 \times 1 \times 9 = 135$.

(B) The given sum is $S = \sum_{r=0}^{20} {}^{50-r}C_{6} = {}^{50}C_{6} + {}^{49}C_{6} + {}^{48}C_{6} + \dots + {}^{30}C_{6}$.
Using the Hockey-stick identity,which states that $\sum_{i=r}^{n} {}^{i}C_{r} = {}^{n+1}C_{r+1}$,we can rewrite the sum.
First,note that ${}^{30}C_{6} = {}^{31}C_{7} - {}^{30}C_{7}$.
Alternatively,we can use the property ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ repeatedly:
$S = {}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{31}C_{6} + {}^{30}C_{6}$.
Adding and subtracting ${}^{30}C_{7}$:
$S = ({}^{50}C_{6} + {}^{49}C_{6} + \dots + {}^{30}C_{6} + {}^{30}C_{7}) - {}^{30}C_{7}$.
Using the identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$ iteratively:
${}^{30}C_{6} + {}^{30}C_{7} = {}^{31}C_{7}$.
${}^{31}C_{6} + {}^{31}C_{7} = {}^{32}C_{7}$.
Continuing this process up to ${}^{50}C_{6} + {}^{50}C_{7} = {}^{51}C_{7}$.
Thus,the sum simplifies to ${}^{51}C_{7} - {}^{30}C_{7}$.

(D) Let the number of questions chosen from sections $A, B,$ and $C$ be $n_1, n_2,$ and $n_3$ respectively,such that $n_1 + n_2 + n_3 = 5$,where $n_i \ge 1$ for $i = 1, 2, 3$.
The possible partitions of $5$ into $3$ parts (each $\ge 1$) are:
$1) (1, 2, 2)$ in any order: $(1, 2, 2), (2, 1, 2), (2, 2, 1)$.
$2) (1, 1, 3)$ in any order: $(1, 1, 3), (1, 3, 1), (3, 1, 1)$.
For case $(1, 2, 2)$:
Number of ways $= \binom{5}{1} \times \binom{5}{2} \times \binom{5}{2} = 5 \times 10 \times 10 = 500$.
Since there are $3$ permutations of $(1, 2, 2)$,total ways $= 3 \times 500 = 1500$.
For case $(1, 1, 3)$:
Number of ways $= \binom{5}{1} \times \binom{5}{1} \times \binom{5}{3} = 5 \times 5 \times 10 = 250$.
Since there are $3$ permutations of $(1, 1, 3)$,total ways $= 3 \times 250 = 750$.
Total number of ways $= 1500 + 750 = 2250$.

(D) The word $'SYLLABUS'$ contains $8$ letters: $S, S, L, L, Y, A, B, U$.
There are $2$ pairs of identical letters ($S, S$ and $L, L$) and $4$ distinct letters $(Y, A, B, U)$.
We need to form a $4$-letter word with $2$ alike and $2$ distinct letters.
Step $1$: Select the pair of identical letters. There are $2$ pairs ($S, S$ or $L, L$),so we can choose $1$ pair in $^2C_1 = 2$ ways.
Step $2$: Select $2$ distinct letters from the remaining $4$ distinct letters $(Y, A, B, U)$. We can choose $2$ letters in $^4C_2 = 6$ ways.
Step $3$: Arrange these $4$ letters (where $2$ are alike and $2$ are distinct). The number of arrangements is $\frac{4!}{2!} = \frac{24}{2} = 12$.
Total number of words = $(\text{Ways to select pair}) \times (\text{Ways to select distinct letters}) \times (\text{Arrangements}) = 2 \times 6 \times 12 = 144$.
Wait,re-evaluating: The distinct letters available are $Y, A, B, U$ and the other pair. If we pick $S, S$,we have $L, Y, A, B, U$ to pick $2$ from. Total distinct letters available are $5$ $(L, Y, A, B, U)$.
Correct calculation: Select $1$ pair from $2$ ($S, S$ or $L, L$) = $2$ ways. Select $2$ distinct letters from the remaining $5$ letters = $^5C_2 = 10$ ways.
Total words = $2 \times 10 \times 12 = 240$.

(C) The word "$LETTER$" contains $6$ letters: $L, E, T, T, E, R$.
The vowels are $E, E$ and the consonants are $L, T, T, R$.
First,arrange the consonants $L, T, T, R$. The number of ways to arrange these $4$ letters (where $T$ repeats twice) is $\frac{4!}{2!} = \frac{24}{2} = 12$.
Now,to ensure vowels never come together,we place them in the gaps created by the consonants:
_ $L$ _ $T$ _ $T$ _ $R$ _
There are $5$ available gaps for the $2$ identical vowels $(E, E)$.
The number of ways to choose $2$ gaps out of $5$ is $^{5}C_{2} = \frac{5 \times 4}{2} = 10$.
Since the vowels are identical,there is only $1$ way to place them in the chosen gaps.
Total number of words = $12 \times 10 = 120$.

(B) There are three families: two families with $3$ members each and one family with $4$ members.
To ensure that members of the same family are not separated,we treat each family as a single unit.
There are $3$ such units (families) to be arranged among themselves,which can be done in $3!$ ways.
Within the first family of $3$ members,they can be arranged in $3!$ ways.
Within the second family of $3$ members,they can be arranged in $3!$ ways.
Within the third family of $4$ members,they can be arranged in $4!$ ways.
Therefore,the total number of ways to seat them such that family members stay together is $3! \cdot 3! \cdot 3! \cdot 4! = (3!)^3 \cdot 4!$.

(C) The word $TABLE$ consists of $5$ distinct letters: $T, A, B, L, E$.
The number of ways to arrange $n$ distinct objects is given by $n!$.
Therefore,the number of ways to arrange the letters of the word $TABLE$ is $5!$.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

(D) The word $MATHEMATICS$ contains $11$ letters in total.
The frequency of each letter is as follows:
$M$ appears $2$ times.
$A$ appears $2$ times.
$T$ appears $2$ times.
$H, E, I, C, S$ appear $1$ time each.
The number of ways to arrange $n$ objects where $n_1, n_2, ...$ are the frequencies of repeating objects is given by the formula $\frac{n!}{n_1! \times n_2! \times ...}$.
Substituting the values: $\frac{11!}{2! \times 2! \times 2!} = \frac{11!}{2 \times 2 \times 2} = \frac{11!}{8}$.

(A) Each of the $6$ letters can be posted in any of the $5$ letter boxes.
Since each letter has $5$ independent choices,the total number of ways to post $6$ letters is $5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^{6}$.

(C) The set of odd digits is ${1, 3, 5, 7, 9}$.
There are $5$ such digits available.
To form a $3$-digit number with distinct digits,we need to select and arrange $3$ digits out of these $5$ available odd digits.
The number of ways to do this is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$,where $n=5$ and $r=3$.
Alternatively,using the fundamental principle of counting:
The first digit can be chosen in $5$ ways.
The second digit can be chosen in $4$ ways (since digits must be distinct).
The third digit can be chosen in $3$ ways.
Total number of $3$-digit numbers $= 5 \times 4 \times 3 = 60$.

(C) The word '$UNIVERSAL$' contains $9$ distinct letters: $U, N, I, V, E, R, S, A, L$.
Since $E, R, S$ must always be together,we treat the group $(ERS)$ as a single unit.
Now,the letters to be arranged are ${U, N, I, V, A, L, (ERS)}$.
This gives us a total of $7$ units to arrange,which can be done in $7!$ ways.
Within the group $(ERS)$,the $3$ letters can be arranged among themselves in $3!$ ways.
Therefore,the total number of arrangements is $7! \times 3!$.
Calculating the values: $7! = 5040$ and $3! = 6$.
Total arrangements $= 5040 \times 6 = 30240$.

(A) The word '$ALGEBRA$' contains $7$ letters,where '$A$' appears $2$ times and all other letters are distinct.
Total number of arrangements $= \frac{7!}{2!} = \frac{5040}{2} = 2520$.
To find the number of arrangements where both '$A$'s come together,we treat the two '$A$'s as a single unit '$AA$'. Now,we have $6$ units to arrange: ${AA, L, G, E, B, R}$.
The number of arrangements where both '$A$'s are together $= 6! = 720$.
The number of ways such that both '$A$'s do not come together = (Total arrangements) - (Arrangements where both '$A$'s are together).
Required number of ways $= 2520 - 720 = 1800$.

(D) The word '$NUTAN$' has $5$ letters: $N, U, T, A, N$.
Consonants are $N, T, N$ ($3$ letters) and vowels are $U, A$ ($2$ letters).
There are $5$ positions in total: $1, 2, 3, 4, 5$.
Odd positions are $1, 3, 5$ ($3$ places) and even positions are $2, 4$ ($2$ places).
Consonants $(N, T, N)$ must occupy the $3$ odd places. The number of ways to arrange them is $\frac{3!}{2!} = 3$ (since $N$ repeats twice).
Vowels $(U, A)$ must occupy the $2$ even places. The number of ways to arrange them is $2! = 2$.
Total number of arrangements $= 3 \times 2 = 6$.

(A) Since the posts are distinct,the order of selection matters.
For the first post,there are $7$ choices.
For the second post,there are $6$ remaining choices.
For the third post,there are $5$ remaining choices.
Therefore,the total number of ways to fill the posts is $7 \times 6 \times 5 = 210$.

(D) To form a $3$-digit number,we have three positions: the hundreds place,the tens place,and the units place.
$1$. The hundreds place cannot be $0$ because a $3$-digit number cannot start with $0$. The available digits are ${2, 3, 5, 7}$. Thus,there are $4$ choices for the hundreds place.
$2$. The tens place can be any of the given digits ${0, 2, 3, 5, 7}$ since repetition is allowed. Thus,there are $5$ choices for the tens place.
$3$. The units place can also be any of the given digits ${0, 2, 3, 5, 7}$ since repetition is allowed. Thus,there are $5$ choices for the units place.
Total number of $3$-digit numbers $= 4 \times 5 \times 5 = 100$.

(A) The $1^{\text{st}}$ item in column $A$ can be matched with any of the $6$ items in column $B$.
The $2^{\text{nd}}$ item in column $A$ can be matched with any of the remaining $5$ items in column $B$.
The $3^{\text{rd}}$ item in column $A$ can be matched with any of the remaining $4$ items in column $B$.
The $4^{\text{th}}$ item in column $A$ can be matched with any of the remaining $3$ items in column $B$.
The $5^{\text{th}}$ item in column $A$ can be matched with any of the remaining $2$ items in column $B$.
The $6^{\text{th}}$ item in column $A$ can be matched with the last remaining $1$ item in column $B$.
Therefore,the total number of possible ways to match the items is given by the permutation of $6$ items,which is $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

(C) $2$-digit number is represented as $XY$,where $X$ is the tens digit and $Y$ is the units digit.
For a number to be even,the units digit $Y$ must be from the set $\{0, 2, 4, 6, 8\}$.
The problem states that the units digit cannot be $0$,so $Y \in \{2, 4, 6, 8\}$. There are $4$ possible choices for the units place.
The tens digit $X$ can be any digit from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Note that $0$ cannot be in the tens place for a $2$-digit number. There are $9$ possible choices for the tens place.
Total number of such even $2$-digit numbers $= (\text{Choices for tens place}) \times (\text{Choices for units place}) = 9 \times 4 = 36$.

(B) $3$-digit number consists of a hundreds place,a tens place,and a units place.
For the hundreds place,the digit $0$ cannot be used,so there are $9$ possible choices $(1, 2, 3, 4, 5, 6, 7, 8, 9)$.
For the tens place,any digit from $0$ to $9$ can be used except the one already placed in the hundreds place. Thus,there are $10 - 1 = 9$ choices.
For the units place,any digit from $0$ to $9$ can be used except the two digits already placed in the hundreds and tens places. Thus,there are $10 - 2 = 8$ choices.
Total number of $3$-digit numbers with no repeated digits $= 9 \times 9 \times 8 = 648$.

(D) $2$-digit number consists of a tens place and a units place.
For the tens place,the possible digits are ${1, 2, 3, 4, 5, 6, 7, 8, 9}$. Thus,there are $9$ choices for the tens place.
For the units place,the digit can be any digit from ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$,but it must be distinct from the digit chosen for the tens place.
Since one digit is already used in the tens place,there are $10 - 1 = 9$ choices remaining for the units place.
Therefore,the total number of $2$-digit numbers with distinct digits is $9 \times 9 = 81$.

(A) To form a $4$-digit number with distinct digits,we have $4$ positions: thousands,hundreds,tens,and units.
$1$. For the thousands place,we cannot use $0$. Therefore,we have $9$ choices $(1, 2, 3, 4, 5, 6, 7, 8, 9)$.
$2$. For the hundreds place,we can use $0$,but we cannot use the digit already placed in the thousands place. Thus,we have $9$ choices ($0$ plus the $8$ remaining digits).
$3$. For the tens place,we cannot use the $2$ digits already placed. Thus,we have $8$ choices.
$4$. For the units place,we cannot use the $3$ digits already placed. Thus,we have $7$ choices.
Total number of $4$-digit numbers with distinct digits $= 9 \times 9 \times 8 \times 7 = 4536$.

(C) The word '$DETAIL$' consists of $6$ letters: $3$ vowels $(A, E, I)$ and $3$ consonants $(D, T, L)$.
There are $6$ positions in total: $1, 2, 3, 4, 5, 6$.
The odd positions are $1, 3,$ and $5$ (total $3$ positions).
The even positions are $2, 4,$ and $6$ (total $3$ positions).
We need to place $3$ vowels in $3$ odd positions,which can be done in $3! = 6$ ways.
We need to place $3$ consonants in $3$ even positions,which can be done in $3! = 6$ ways.
Therefore,the total number of arrangements is $6 \times 6 = 36$.

(B) To form a $3$-digit even number,the unit's place must be occupied by an even digit ($0$ or $2$).
Case $1$: Unit's place is $0$.
The remaining two places (hundreds and tens) can be filled by the remaining $4$ digits $(2, 3, 5, 7)$.
The number of ways to fill the hundreds place is $4$ and the tens place is $3$.
Total ways for Case $1 = 4 \times 3 = 12$.
Case $2$: Unit's place is $2$.
The hundreds place cannot be $0$,and it cannot be $2$ (since repetition is not allowed).
So,the hundreds place can be filled by any of the remaining $3$ digits $(3, 5, 7)$.
The tens place can then be filled by any of the remaining $3$ digits (including $0$ but excluding the digits used in the hundreds and unit's places).
Total ways for Case $2 = 3 \times 3 = 9$.
Total even numbers $= 12 + 9 = 21$.

(C) The first $3$ questions each have $4$ choices. The number of ways to answer these is $4 \times 4 \times 4 = 4^3 = 64$.
The next $3$ questions each have $5$ choices. The number of ways to answer these is $5 \times 5 \times 5 = 5^3 = 125$.
According to the fundamental principle of counting,the total number of sequences of answers is the product of the number of ways to answer each set of questions.
Total sequences = $64 \times 125 = 8000$.

(D) Total number of persons $= 7 + 6 = 13$.
We need to select $5$ persons such that at least $3$ are men.
This can be done in the following cases:
Case $1$: $3$ men and $2$ women.
Number of ways $= {}^{7}C_{3} \times {}^{6}C_{2} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{6 \times 5}{2 \times 1} = 35 \times 15 = 525$.
Case $2$: $4$ men and $1$ woman.
Number of ways $= {}^{7}C_{4} \times {}^{6}C_{1} = {}^{7}C_{3} \times 6 = 35 \times 6 = 210$.
Case $3$: $5$ men and $0$ women.
Number of ways $= {}^{7}C_{5} \times {}^{6}C_{0} = {}^{7}C_{2} \times 1 = \frac{7 \times 6}{2 \times 1} = 21$.
Total number of ways $= 525 + 210 + 21 = 756$.

(C) The word '$LEADING$' consists of $7$ distinct letters: $L, E, A, D, I, N, G$.
The vowels in the word are $E, A, I$ (total $3$ vowels).
To ensure the vowels always come together,we treat the group $(E, A, I)$ as a single unit or entity.
Now,we have the letters ${L, D, N, G}$ and the single unit $(E, A, I)$.
This gives us a total of $4 + 1 = 5$ entities to arrange.
These $5$ entities can be arranged in $5!$ ways.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Within the group $(E, A, I)$,the $3$ vowels can be arranged among themselves in $3!$ ways.
$3! = 3 \times 2 \times 1 = 6$.
Therefore,the total number of arrangements is $5! \times 3! = 120 \times 6 = 720$.

(C) Step $1$: Select $3$ consonants out of $7$. The number of ways is ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Step $2$: Select $2$ vowels out of $4$. The number of ways is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
Step $3$: The total number of letters selected is $3 + 2 = 5$. These $5$ letters can be arranged among themselves in $5!$ ways.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Step $4$: The total number of words that can be formed is the product of the number of selections and the number of arrangements.
Total words $= 35 \times 6 \times 120 = 210 \times 120 = 25200$.

(D) Total number of children $= 6 + 4 = 10$.
Total number of ways to select $4$ children from $10$ is given by ${}^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
Number of ways to select $0$ boys (i.e.,all $4$ are girls) is given by ${}^{4}C_{4} = 1$.
To find the number of ways to select at least one boy,we subtract the number of ways to select no boys from the total number of ways.
Number of ways $= {}^{10}C_{4} - {}^{4}C_{4} = 210 - 1 = 209$.

(D) To form a $3$-digit number divisible by $5$,the digit at the units place must be $5$.
Since the digits cannot be repeated,we have $1$ choice for the units place (which is $5$).
For the remaining $2$ positions (tens and hundreds),we have $5$ remaining digits available $(2, 3, 6, 7, 9)$.
The number of ways to fill the hundreds place is $5$.
The number of ways to fill the tens place is $4$.
Total number of $3$-digit numbers $= 5 \times 4 \times 1 = 20$.

(C) The number of ways to select $5$ men from $8$ men is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
${}^{8}C_{5} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
The number of ways to select $6$ women from $10$ women is ${}^{10}C_{6}$.
${}^{10}C_{6} = {}^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
To find the total number of ways to form the committee,we multiply the number of ways to select the men and the women:
Total ways $= 56 \times 210 = 11760$.

(A) The word $MEDITERRANEAN$ consists of $13$ letters: $M, E, D, I, T, E, R, R, A, N, E, A, N$. The frequency of letters is: $E:3, R:2, A:2, N:2, M:1, D:1, I:1, T:1$. Total distinct letters available are ${M, E, D, I, T, R, A, N}$,which is $8$ distinct types.
We need to form a $4$-letter word where the first letter is $E$ and the last is $R$. The word structure is $E . . R$.
We need to fill the $2$ middle positions using the remaining letters. The available letters after fixing $E$ and $R$ are: $E:2, R:1, A:2, N:2, M:1, D:1, I:1, T:1$. Total $11$ letters remain.
Case $1$: The two middle letters are the same. The letters that appear at least twice are $E, A, N$. We can choose one of these $3$ pairs. Number of ways $= 3$.
Case $2$: The two middle letters are different. We need to choose $2$ distinct letters from the $8$ available types ${M, E, D, I, T, R, A, N}$. The number of ways to choose and arrange $2$ distinct letters is $P(8, 2) = 8 \times 7 = 56$.
Total number of words $= 56 + 3 = 59$.

(A) The word '$MATHEMATICS$' consists of $11$ letters: $M, A, T, H, E, M, A, T, I, C, S$.
The vowels are $A, A, E, I$ (total $4$ vowels).
The consonants are $M, T, H, M, T, C, S$ (total $7$ consonants).
Treating the $4$ vowels as a single unit,we have $7$ consonants + $1$ unit of vowels = $8$ entities to arrange.
Among these $8$ entities,the letter $M$ repeats $2$ times and $T$ repeats $2$ times.
The number of ways to arrange these $8$ entities is $\frac{8!}{2! \times 2!} = \frac{40320}{4} = 10080$.
Within the vowel unit,the $4$ vowels $(A, A, E, I)$ can be arranged in $\frac{4!}{2!}$ ways (since $A$ repeats $2$ times).
$\frac{4!}{2!} = \frac{24}{2} = 12$.
Total number of arrangements = $10080 \times 12 = 120960$.

(B) To select $2$ men out of $6$,the number of ways is given by the combination formula $nC_r = \frac{n!}{r!(n-r)!}$.
Number of ways to select $2$ men out of $6 = 6C_2 = \frac{6 \times 5}{2 \times 1} = 15$.
Number of ways to select $3$ women out of $8 = 8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Since the selection of men and women are independent events,the total number of ways to form the committee is the product of these two values.
Total number of ways = $15 \times 56 = 840$.

(A) To form a committee of $5$ people consisting only of men,we need to select $5$ men from the available $6$ men.
The number of ways to choose $r$ items from $n$ items is given by the combination formula: ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Here,$n = 6$ and $r = 5$.
Number of ways = ${}^{6}C_{5} = \frac{6!}{5!(6-5)!} = \frac{6!}{5! \times 1!} = \frac{6 \times 5!}{5! \times 1} = 6$.
Therefore,there are $6$ ways to select the committee.

(D) Total number of people $= 6 + 8 = 14$.
Total number of ways to select $5$ people out of $14$ is given by ${}^{14}C_{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$.
Number of ways such that no woman is selected means all $5$ people are selected from the $6$ men. This is given by ${}^{6}C_{5} = 6$.
The number of ways to select at least one woman is equal to (Total ways) - (Ways with no women).
Required number of ways $= 2002 - 6 = 1996$.

(C) At most one man means either no man is selected or exactly one man is selected.
Case $1$: No man is selected.
This means all $5$ members are selected from $8$ women.
Number of ways $= {}^{8}C_{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Case $2$: Exactly one man is selected.
This means $1$ man is selected from $6$ men and $4$ women are selected from $8$ women.
Number of ways $= {}^{6}C_{1} \times {}^{8}C_{4} = 6 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 6 \times 70 = 420$.
Total number of ways $= 56 + 420 = 476$.

(C) To form a four-digit even number,the units place must be filled by $0, 2, 4,$ or $6$.
Case $I$: When $0$ is at the units place.
The units place can be filled in $1$ way. The remaining $3$ places can be filled using the remaining $6$ digits $(1, 2, 3, 4, 5, 6)$ in $^6P_3$ ways.
Number of ways $= 1 \times ^6P_3 = \frac{6!}{3!} = 6 \times 5 \times 4 = 120$.
Case $II$: When $0$ is not at the units place.
The units place can be filled by $2, 4,$ or $6$ in $3$ ways. The thousands place cannot be $0$ and cannot be the digit used in the units place,so it can be filled in $5$ ways (out of $6$ remaining digits). The remaining $2$ places can be filled by the remaining $5$ digits in $^5P_2$ ways.
Number of ways $= 3 \times 5 \times ^5P_2 = 15 \times \frac{5!}{3!} = 15 \times 20 = 300$.
Total even numbers $= 120 + 300 = 420$.

(B) To find the number of different sums of money that can be formed using four distinct notes,we consider all possible non-empty combinations of these notes.
Since each note is distinct,choosing any subset of the four notes will result in a unique sum.
The total number of non-empty subsets of a set of $4$ elements is given by $2^n - 1$,where $n = 4$.
Number of different sums $= 2^4 - 1 = 16 - 1 = 15$.
Alternatively,this can be calculated as the sum of combinations of taking $1, 2, 3,$ or $4$ notes at a time:
Number of sums $= {^4C_1} + {^4C_2} + {^4C_3} + {^4C_4} = 4 + 6 + 4 + 1 = 15$.

(A) The total number of ways to arrange $12$ persons around a round table is given by $(n-1)!$,where $n=12$.
Total arrangements $= (12-1)! = 11!$.
Next,we calculate the number of ways where $2$ particular persons sit side by side. We treat these $2$ persons as a single unit. Now,we have $11$ units to arrange around a round table,which can be done in $(11-1)! = 10!$ ways.
Since the $2$ persons within the unit can be arranged in $2!$ ways,the total number of ways where they sit together is $10! \times 2!$.
Finally,the number of arrangements where the $2$ particular persons do not sit side by side is the total arrangements minus the arrangements where they sit together:
$= 11! - (10! \times 2!)$
$= (11 \times 10!) - (2 \times 10!)$
$= (11-2) \times 10!$
$= 9 \times 10!$

(D) The word $ARTICLE$ contains $7$ letters in total.
The vowels in the word are $A, I, E$,which are $3$ in number.
The consonants in the word are $R, T, C, L$,which are $4$ in number.
In a $7$-letter word,the positions are $1, 2, 3, 4, 5, 6, 7$. The even positions are $2, 4, 6$,which are $3$ in total.
Since the $3$ vowels must occupy the $3$ even places,the number of ways to arrange them is $^3P_3 = 3! = 6$.
The remaining $4$ consonants must occupy the remaining $4$ odd positions $(1, 3, 5, 7)$. The number of ways to arrange them is $^4P_4 = 4! = 24$.
Therefore,the total number of words that can be formed is $3! \times 4! = 6 \times 24 = 144$.

(C) We need to form $5$-digit numbers greater than $24,000$ using the digits ${1, 2, 3, 4, 5}$ without repetition.
Case $1$: The first digit is $2$. To be greater than $24,000$,the second digit must be $4$ or $5$.
- If the second digit is $4$,the remaining $3$ positions can be filled by the remaining $3$ digits in $3! = 6$ ways.
- If the second digit is $5$,the remaining $3$ positions can be filled by the remaining $3$ digits in $3! = 6$ ways.
Total for Case $1 = 6 + 6 = 12$ numbers.
Case $2$: The first digit is $3, 4,$ or $5$ ($3$ choices).
The remaining $4$ positions can be filled by the remaining $4$ digits in $4! = 24$ ways.
Total for Case $2 = 3 \times 24 = 72$ numbers.
Total numbers $= 12 + 72 = 84$.

(D) Numbers greater than $100$ can be either $3$-digit or $4$-digit numbers.
For a number to be divisible by $5$,the digit $5$ must be at the units place.
Case $I$: $3$-digit numbers.
The units place is fixed as $5$. The remaining $2$ places can be filled by the remaining $3$ digits $(3, 4, 6)$ in $P(3, 2) = 3 \times 2 = 6$ ways.
Case $II$: $4$-digit numbers.
The units place is fixed as $5$. The remaining $3$ places can be filled by the remaining $3$ digits $(3, 4, 6)$ in $P(3, 3) = 3 \times 2 \times 1 = 6$ ways.
Total number of required numbers $= 6 + 6 = 12$.

(C) number is divisible by $4$ if its last two digits form a number divisible by $4$.
Given digits are ${1, 2, 3, 4, 5, 6}$.
The possible $2$-digit numbers formed by these digits that are divisible by $4$ are: $12, 16, 24, 32, 36, 52, 56, 64$.
There are $8$ such possible pairs for the last two digits.
For each pair,we need to fill the remaining $3$ positions using the remaining $4$ digits from the set ${1, 2, 3, 4, 5, 6}$.
The number of ways to choose and arrange $3$ digits out of $4$ is given by $P(4, 3) = 4 \times 3 \times 2 = 24$.
Since there are $8$ such pairs,the total number of $5$-digit numbers $n$ is $8 \times 24 = 192$.

(D) chess board has $32$ black and $32$ white squares.
Total ways to choose one white and one black square $= {}^{32}C_{1} \times {}^{32}C_{1} = 32 \times 32 = 1024$.
There are $8$ rows,and each row contains $4$ white squares and $4$ black squares.
Ways to choose a white and a black square in the same row $= 8 \times ({}^{4}C_{1} \times {}^{4}C_{1}) = 8 \times 16 = 128$.
Similarly,there are $8$ columns,and each column contains $4$ white squares and $4$ black squares.
Ways to choose a white and a black square in the same column $= 8 \times ({}^{4}C_{1} \times {}^{4}C_{1}) = 8 \times 16 = 128$.
Total ways where squares lie in the same row or column $= 128 + 128 = 256$.
Therefore,the required number of ways $= 1024 - 256 = 768$.

(B) To ensure that we have at least two balls of the same colour,we apply the Pigeonhole Principle.
If we draw $1$ ball,it can be either black or white.
If we draw $2$ balls,they could be one black and one white (different colours).
If we draw $3$ balls,according to the Pigeonhole Principle,since there are only $2$ colours available,at least $2$ balls must be of the same colour.
Therefore,the minimum number of balls required is $3$.

(C) The total number of available digits is $n = 5$ $(1, 2, 3, 4, 5)$.
We need to form a number with $r = 4$ digits.
Since the order of digits matters in forming a number,we use the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.
Substituting the values: $P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$.
Therefore,$120$ four-digit numbers can be formed.

(D) Each question has $2$ possible outcomes: True $(T)$ or False $(F)$.
Since there are $10$ questions and each is answered independently,the total number of possible sequences is calculated by the multiplication principle.
Total number of sequences $= 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$.
$2^{10} = 1024$.

(D) To find the number of handshakes,we need to choose $2$ persons out of $10$ to perform a handshake.
This is a combination problem because the order of the two people shaking hands does not matter.
The formula for combinations is given by $^nC_r = \frac{n!}{r!(n-r)!}$.
Here,$n = 10$ and $r = 2$.
Number of ways $= ^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
Therefore,there are $45$ possible ways for ten persons to shake hands with one another.

(B) To ensure that $A$ and $B$ are always together,we treat them as a single unit or block $(AB)$.
Now,we have this block $(AB)$ and the remaining two children,making a total of $3$ entities to arrange.
The number of ways to arrange these $3$ entities in a line is $3! = 3 \times 2 \times 1 = 6$.
Within the block $(AB)$,the two children $A$ and $B$ can arrange themselves in $2! = 2 \times 1 = 2$ ways (i.e.,$AB$ or $BA$).
Therefore,the total number of ways is $3! \times 2! = 6 \times 2 = 12$.

(D) The number of ways in which $n$ distinct objects can be arranged in a line is given by $n!$.
Here,the number of persons $n = 6$.
Therefore,the total number of ways to arrange them in a queue is $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

(B) $3$-digit number consists of three places: Hundreds,Tens,and Units.
For the Hundreds place,the digit '$0$' cannot be used because a $3$-digit number cannot start with $0$. Thus,we can choose from ${1, 3, 5, 7}$,which gives $4$ options.
For the Tens place,since repetition is allowed,all $5$ digits ${0, 1, 3, 5, 7}$ can be used,giving $5$ options.
For the Units place,since repetition is allowed,all $5$ digits ${0, 1, 3, 5, 7}$ can be used,giving $5$ options.
Therefore,the total number of $3$-digit numbers $= 4 \times 5 \times 5 = 100$.