Definition of permutation, Number of permutations with or without repetition, Conditional permutations Questions in English

(B) The $1^{st}$ ring has $10$ digits $(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$.
The $2^{nd}$ ring has prime numbers greater than $2$ and less than $30$: ${3, 5, 7, 11, 13, 17, 19, 23, 29}$. There are $9$ such numbers.
The $3^{rd}$ ring has all vowels: ${a, e, i, o, u}$. There are $5$ such vowels.
Total number of ways to set the lock $= 10 \times 9 \times 5 = 450$.
Since there is only $1$ correct combination to open the lock,the total number of unsuccessful attempts $= 450 - 1 = 449$.

(B) The total number of permutations of digits $1, 2, 3, 4$ is $4! = 24$.
We want to exclude cases where $n+1$ immediately follows $n$. These are the patterns $12, 23, 34$.
Let $S$ be the set of all permutations. Let $A$ be the set of permutations containing $12$,$B$ containing $23$,and $C$ containing $34$.
We want to find $24 - |A \cup B \cup C|$.
Using the Principle of Inclusion-Exclusion:
$|A| = |B| = |C| = 3! = 6$.
$|A \cap B|$ (contains $123$) = $2! = 2$.
$|B \cap C|$ (contains $234$) = $2! = 2$.
$|A \cap C|$ (contains $12$ and $34$) = $2! = 2$.
$|A \cap B \cap C|$ (contains $1234$) = $1! = 1$.
$|A \cup B \cup C| = (6+6+6) - (2+2+2) + 1 = 18 - 6 + 1 = 13$.
Total valid numbers = $24 - 13 = 11$.

(D) Each of the $n$ places in the $n$-digit number can be filled in $3$ ways (using digits $2, 5,$ or $7$).
Therefore,the total number of distinct $n$-digit numbers that can be formed is $3^n$.
We need to find the smallest integer $n$ such that $3^n \geq 900$.
Calculating powers of $3$:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
$3^5 = 243$
$3^6 = 729$
$3^7 = 2187$
Since $3^6 = 729 < 900$ and $3^7 = 2187 \geq 900$,the smallest value of $n$ is $7$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$.
The thousands place can be filled with $2, 3,$ or $4$ because the number must be between $2,000$ and $5,000$.
Case $1$: Thousands place is $2$.
Possible sets of remaining $3$ digits from ${0, 1, 3, 4}$ such that the sum is divisible by $3$:
- ${0, 1, 3}$ (sum $= 2+0+1+3 = 6$)
- ${0, 3, 4}$ (sum $= 2+0+3+4 = 9$)
Each set can be arranged in $3! = 6$ ways. Total $= 2 \times 6 = 12$.
Case $2$: Thousands place is $3$.
Possible sets of remaining $3$ digits from ${0, 1, 2, 4}$ such that the sum is divisible by $3$:
- ${0, 1, 2}$ (sum $= 3+0+1+2 = 6$)
- ${0, 2, 4}$ (sum $= 3+0+2+4 = 9$)
Each set can be arranged in $3! = 6$ ways. Total $= 2 \times 6 = 12$.
Case $3$: Thousands place is $4$.
Possible sets of remaining $3$ digits from ${0, 1, 2, 3}$ such that the sum is divisible by $3$:
- ${0, 2, 1}$ (sum $= 4+0+2+1 = 7$ - No)
- ${0, 2, 3}$ (sum $= 4+0+2+3 = 9$)
- ${1, 2, 3}$ (sum $= 4+1+2+3 = 10$ - No)
Only ${0, 2, 3}$ works. Total $= 1 \times 6 = 6$.
Total numbers $= 12 + 12 + 6 = 30$.

(C) The letters in the word $QUEEN$ are $E, E, N, Q, U$. Arranging them in alphabetical order: $E, E, N, Q, U$.
$(i)$ Words starting with $E$: The remaining letters are $E, N, Q, U$ (all distinct). Number of words $= 4! = 24$.
$(ii)$ Words starting with $N$: The remaining letters are $E, E, Q, U$. Number of words $= \frac{4!}{2!} = 12$.
$(iii)$ Words starting with $QE$: The remaining letters are $E, N, U$. Number of words $= 3! = 6$.
$(iv)$ Words starting with $QN$: The remaining letters are $E, E, U$. Number of words $= \frac{3!}{2!} = 3$.
$(v)$ The next word is $QUEEN$ itself,which is the $1^{st}$ word after the above arrangements.
Therefore,the rank of the word $QUEEN = 24 + 12 + 6 + 3 + 1 = 46^{th}$.

(B) The word $MEDITERRANEAN$ contains $13$ letters: $M(1), E(3), D(1), I(1), T(1), R(2), A(2), N(2)$.
We need to form a $4$-letter word of the form $R \_ \_ E$.
The two middle positions can be filled in two ways:
$1$. Both letters are identical: The available pairs are $(E, E), (A, A), (N, N)$. There are $3$ such pairs.
$2$. Both letters are distinct: We choose $2$ distinct letters from the set ${M, E, D, I, T, R, A, N}$. There are $8$ distinct letters available. The number of ways to arrange $2$ distinct letters in the $2$ middle positions is $^8P_2 = 8 \times 7 = 56$.
Total number of words = $3 56 = 59$.

(B) To form a $4-$ digit number using the digits $3, 4, 5,$ and $6$ without repetition,we have $4! = 24$ total numbers.
If we fix a digit at the unit's place,the remaining $3$ positions can be filled by the remaining $3$ digits in $3! = 6$ ways.
Therefore,each of the digits $3, 4, 5,$ and $6$ appears in the unit's place exactly $6$ times.
The sum of the digits in the unit's place is:
$= (6 \times 3) + (6 \times 4) + (6 \times 5) + (6 \times 6)$
$= 6 \times (3 + 4 + 5 + 6)$
$= 6 \times 18$
$= 108$

(B) The total number of digits is $8$. The odd places are $1, 3, 5, 7$ ($4$ places) and the even places are $2, 4, 6, 8$ ($4$ places).
The given digits are $1, 1, 2, 2, 2, 3, 4, 4$. The odd digits are $1, 1, 3$ (total $3$ digits) and the even digits are $2, 2, 2, 4, 4$ (total $5$ digits).
We are given that odd digits do not occupy odd places. This means the $3$ odd digits must be placed in the $4$ available even places.
The number of ways to arrange the $3$ odd digits $(1, 1, 3)$ in $4$ even places is given by $\frac{4!}{2!} = 12$.
After placing the $3$ odd digits,we have $5$ remaining places (the $1$ remaining even place and the $4$ odd places) to fill with the remaining $5$ even digits $(2, 2, 2, 4, 4)$.
The number of ways to arrange these $5$ digits is $\frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10$.
Therefore,the total number of such $8$-digit numbers is $12 \times 10 = 120$.

(D) The set of digits is $\{2, 3, 5, 7, 9\}$. The total number of $5$-digit numbers formed without repetition is $5! = 120$.
For $p$: The numbers must exceed $20000$. Since all given digits are $\ge 2$,any $5$-digit number formed using these digits will be $\ge 23579$,which is $> 20000$. Thus,$p = 5! = 120$.
For $q$: The numbers must lie between $30000$ and $90000$. This means the first digit must be $3, 5,$ or $7$.
There are $3$ choices for the first digit. The remaining $4$ positions can be filled by the remaining $4$ digits in $4!$ ways.
So,$q = 3 \times 4! = 3 \times 24 = 72$.
The ratio $p : q = 120 : 72$.
Dividing both by $24$,we get $120/24 : 72/24 = 5 : 3$.

(A) The set of letters is $\{a, b, c, d, e, f\}$,which contains $2$ vowels $(\{a, e\})$ and $4$ consonants $(\{b, c, d, f\})$.
We need to form arrangements of $3$ letters containing at least one vowel.
Total arrangements with at least one vowel = (Total arrangements of $3$ letters) - (Arrangements with no vowels).
Total arrangements of $3$ letters from $6$ distinct letters = $^6P_3 = 6 \times 5 \times 4 = 120$.
Arrangements with no vowels (only consonants) = $^4P_3 = 4 \times 3 \times 2 = 24$.
Therefore,the number of arrangements with at least one vowel = $120 - 24 = 96$.

(B) We need to find the number of natural numbers less than $7,000$ using the digits ${0, 1, 3, 7, 9}$ with repetition allowed.
Case $1$: $1$-digit,$2$-digit,or $3$-digit numbers.
For a $1$-digit number,there are $4$ choices (excluding $0$): ${1, 3, 7, 9}$.
For a $2$-digit number,the first digit has $4$ choices and the second has $5$ choices: $4 \times 5 = 20$.
For a $3$-digit number,the first digit has $4$ choices and the next two have $5$ choices each: $4 \times 5 \times 5 = 100$.
Total for $1, 2, 3$ digits $= 4 + 20 + 100 = 124$.
Case $2$: $4$-digit numbers less than $7,000$.
The first digit can be $1$ or $3$ ($2$ choices).
The remaining three positions can each be filled by any of the $5$ digits ($5 \times 5 \times 5 = 125$ choices).
Total $4$-digit numbers $= 2 \times 125 = 250$.
Total count $= 124 + 250 = 374$.

(A) The total number of digits is $9$. The digits are $1, 1, 2, 2, 2, 2, 3, 4, 4$.
Odd digits are $1, 1, 3$ (total $3$ odd digits).
Even digits are $2, 2, 2, 2, 4, 4$ (total $6$ even digits).
There are $4$ even places $(2^{nd}, 4^{th}, 6^{th}, 8^{th})$ and $5$ odd places $(1^{st}, 3^{rd}, 5^{th}, 7^{th}, 9^{th})$.
We need to place $3$ odd digits in $4$ even places. This is not possible as we have $3$ odd digits and $4$ even places. However,the question implies we must place the $3$ odd digits into the $4$ even places. Since we have $3$ odd digits $(1, 1, 3)$,we choose $3$ places out of $4$ even places in $^4C_3$ ways.
The number of ways to arrange $1, 1, 3$ in these $3$ places is $\frac{3!}{2!} = 3$.
The remaining $6$ digits $(2, 2, 2, 2, 4, 4)$ must be placed in the remaining $6$ places. The number of ways to arrange them is $\frac{6!}{4!2!} = 15$.
Total numbers $= ^4C_3 \times \frac{3!}{2!} \times \frac{6!}{4!2!} = 4 \times 3 \times 15 = 180$.

(C) We need to find the number of four-digit numbers greater than $4321$ using digits $\{0, 1, 2, 3, 4, 5\}$.
Case $1$: Numbers starting with $5$:
The first digit is $5$. The remaining $3$ positions can be filled by any of the $6$ digits $(0-5)$.
Number of ways $= 1 \times 6 \times 6 \times 6 = 216$.
Case $2$: Numbers starting with $44$ or $45$:
The first digit is $4$. The second digit is $4$ or $5$ ($2$ choices). The remaining $2$ positions can be filled by any of the $6$ digits.
Number of ways $= 1 \times 2 \times 6 \times 6 = 72$.
Case $3$: Numbers starting with $43$:
The first two digits are $43$. The third digit must be greater than $2$,so it can be $3, 4, 5$ ($3$ choices).
The fourth digit can be any of the $6$ digits.
Number of ways $= 1 \times 1 \times 3 \times 6 = 18$.
Case $4$: Numbers starting with $432$:
The first three digits are $432$. The fourth digit must be greater than $1$,so it can be $2, 3, 4, 5$ ($4$ choices).
Number of ways $= 1 \times 1 \times 1 \times 4 = 4$.
Total count $= 216 + 72 + 18 + 4 = 310$.

(A) To form a $6$-digit number using all five digits $\{1, 3, 5, 7, 9\}$,one digit must be repeated exactly twice,and the other four digits must appear exactly once.
Step $1$: Select the digit to be repeated from the $5$ available digits. This can be done in $^5C_1 = 5$ ways.
Step $2$: Arrange these $6$ digits (where one digit is repeated twice) in a sequence. The number of permutations is given by $\frac{6!}{2!}$.
Step $3$: The total number of such $6$-digit numbers is $^5C_1 \times \frac{6!}{2!} = 5 \times \frac{720}{2} = 5 \times 360 = 1800$.
Note: $\frac{5}{2}(6!) = 2.5 \times 720 = 1800$. Thus,the correct option is $A$.

(A) five-digit number is represented as $\_ \;\_\;\_\;\underline{2}\;\_$.
The $10^{\text{th}}$ place is fixed as $2$.
For the $10,000^{\text{th}}$ place,we cannot use $0$ or $2$,so there are $8$ choices $(1, 3, 4, 5, 6, 7, 8, 9)$.
For the $1,000^{\text{th}}$ place,we can use $0$ and any of the remaining $7$ digits,so there are $8$ choices.
For the $100^{\text{th}}$ place,there are $7$ choices remaining.
For the units place,there are $6$ choices remaining.
Total number of such five-digit numbers $= 8 \times 8 \times 7 \times 6 = 2688$.
Given that the number is $336k$,we have $336k = 2688$.
Therefore,$k = \frac{2688}{336} = 8$.

(A) The word $ROSE$ contains $4$ distinct letters: $R, O, S, E$.
To form a $4$ letter word without repetition,we need to arrange these $4$ letters in $4$ vacant places.
The first place can be filled in $4$ ways.
The second place can be filled in $3$ ways.
The third place can be filled in $2$ ways.
The fourth place can be filled in $1$ way.
By the multiplication principle,the total number of words is $4 \times 3 \times 2 \times 1 = 24$.

(A) There will be as many signals as there are ways of filling in $2$ vacant places in succession by the $4$ flags of different colours.
The upper vacant place can be filled in $4$ different ways by any one of the $4$ flags.
Following which,the lower vacant place can be filled in $3$ different ways by any one of the remaining $3$ different flags.
Hence,by the multiplication principle,the required number of signals $= 4 \times 3 = 12$.

(A) $2$ digit number has a ten's place and a unit's place.
For the number to be even,the unit's place must be filled by an even digit from the set $\{1, 2, 3, 4, 5\}$.
The available even digits are $2$ and $4$. Thus,the unit's place can be filled in $2$ ways.
Since repetition of digits is allowed,the ten's place can be filled by any of the $5$ given digits $(1, 2, 3, 4, 5)$. Thus,the ten's place can be filled in $5$ ways.
By the multiplication principle,the total number of $2$ digit even numbers is $5 \times 2 = 10$.

(A) There are $3$ vacant places to be filled by the given $5$ digits $(1, 2, 3, 4, 5)$.
Since repetition is allowed,each of the $3$ places can be filled in $5$ different ways.
By the multiplication principle,the total number of $3$-digit numbers is $5 \times 5 \times 5 = 125$.

(A) To form a $3$-digit number using the digits $1, 2, 3, 4,$ and $5$ without repetition,we need to fill $3$ places (hundreds,tens,and units).
The hundreds place can be filled in $5$ ways (using any of the digits $1, 2, 3, 4, 5$).
Since repetition is not allowed,the tens place can be filled in $4$ remaining ways.
The units place can then be filled in $3$ remaining ways.
By the multiplication principle,the total number of $3$-digit numbers is $5 \times 4 \times 3 = 60$.

(A) To form a $3$-digit even number,the units place must be filled by an even digit from the given set $\{1, 2, 3, 4, 5, 6\}$.
The even digits available are $2, 4, 6$. Thus,the units place can be filled in $3$ ways.
Since repetition is allowed,the tens place can be filled by any of the $6$ digits in $6$ ways.
Similarly,the hundreds place can be filled by any of the $6$ digits in $6$ ways.
By the multiplication principle,the total number of $3$-digit even numbers is $6 \times 6 \times 3 = 108$.

(A) The number of ways to form a $4$-letter code is equivalent to the number of ways to fill $4$ vacant places in succession using the first $10$ letters of the English alphabet,where repetition is not allowed.
The first place can be filled in $10$ ways.
The second place can be filled in $9$ ways.
The third place can be filled in $8$ ways.
The fourth place can be filled in $7$ ways.
By the multiplication principle,the total number of ways is $10 \times 9 \times 8 \times 7 = 5040$.
Thus,$5040$ four-letter codes can be formed.

(A) $5$-digit telephone number starts with $67$.
This means the first two places are fixed as $6$ and $7$.
Since no digit can be repeated,we have $10 - 2 = 8$ digits remaining to fill the remaining $3$ places.
The third place can be filled in $8$ ways.
The fourth place can be filled in $7$ ways.
The fifth place can be filled in $6$ ways.
By the multiplication principle,the total number of such telephone numbers is $8 \times 7 \times 6 = 336$.

(B) When a coin is tossed once,the number of outcomes is $2$ (Head and Tail).
Since the coin is tossed $3$ times,by the multiplication principle,the total number of possible outcomes is $2 \times 2 \times 2 = 8$.

(A) Each signal requires the use of $2$ flags,one below the other.
The number of signals is equal to the number of ways of filling $2$ vacant places in succession using $5$ flags of different colours.
The upper vacant place can be filled in $5$ different ways by any one of the $5$ flags.
After filling the upper place,the lower vacant place can be filled in $4$ different ways by any one of the remaining $4$ flags.
By the multiplication principle,the total number of different signals that can be generated is $5 \times 4 = 20$.

(A) The factorial of a number $n$,denoted by $n!$,is the product of all positive integers from $1$ to $n$.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

(A) The factorial of a number $n$,denoted by $n!$,is the product of all positive integers from $1$ to $n$.
$7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040$

(A) We know that $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Therefore,$7! - 5! = 5040 - 120 = 4920$.

(A) We have $\frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!}$.
By canceling $5!$ from the numerator and the denominator,we get $7 \times 6$.
Therefore,$\frac{7!}{5!} = 42$.

(A) The factorial of a number $n$ is defined as $n! = 1 \times 2 \times 3 \times \dots \times n$.
For $n = 8$,we have:
$8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 = 40320$.

(A) We know that the factorial of a number $n$ is defined as $n! = 1 \times 2 \times 3 \times \dots \times n$.
First,calculate $4!$:
$4! = 1 \times 2 \times 3 \times 4 = 24$.
Next,calculate $3!$:
$3! = 1 \times 2 \times 3 = 6$.
Finally,subtract the values:
$4! - 3! = 24 - 6 = 18$.

(B) $3! = 1 \times 2 \times 3 = 6$
$4! = 1 \times 2 \times 3 \times 4 = 24$
$\therefore 3! + 4! = 6 + 24 = 30$
$7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040$
$\therefore 3! + 4! \neq 7!$

(B) Given the expression $\frac{n!}{(n-r)!}$.
Substitute $n=6$ and $r=2$ into the expression:
$\frac{6!}{(6-2)!} = \frac{6!}{4!}$
$= \frac{6 \times 5 \times 4!}{4!}$
$= 6 \times 5 = 30$.

(A) Given $n=9$ and $r=5$.
Substituting these values into the expression $\frac{n!}{(n-r)!}$:
$\frac{9!}{(9-5)!} = \frac{9!}{4!}$
$= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$
$= 9 \times 8 \times 7 \times 6 \times 5$
$= 15120$.

(A) The word $ALLAHABAD$ contains $9$ letters in total.
These letters are: $A, L, L, A, H, A, B, A, D$.
The frequency of each letter is:
$A: 4$
$L: 2$
$H: 1$
$B: 1$
$D: 1$
Using the formula for permutations of a multiset,the number of arrangements is given by $\frac{n!}{n_1! n_2! ... n_k!}$.
Here,$n = 9$,$n_1 = 4$ (for $A$),and $n_2 = 2$ (for $L$).
Number of permutations $= \frac{9!}{4! 2!} = \frac{362880}{24 \times 2} = \frac{362880}{48} = 7560$.

(A) Since the order of digits matters,we are looking for the number of permutations of $9$ distinct digits taken $4$ at a time.
The number of ways to form a $4$-digit number is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 9$ and $r = 4$.
Therefore,the number of $4$-digit numbers $= ^9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$.

(A) Every number between $100$ and $1000$ is a $3$-digit number.
To form a $3$-digit number,the hundreds place cannot be $0$.
The hundreds place can be filled by any of the $5$ digits $(1, 2, 3, 4, 5)$,so there are $5$ choices.
The tens place can be filled by any of the remaining $5$ digits (including $0$),so there are $5$ choices.
The units place can be filled by any of the remaining $4$ digits,so there are $4$ choices.
Therefore,the total number of such $3$-digit numbers is $5 \times 5 \times 4 = 100$.

(A) Given the equation: $^{n}P_{5} = 42 \, ^{n}P_{3}$.
Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:
$n(n-1)(n-2)(n-3)(n-4) = 42 \, n(n-1)(n-2)$.
Since $n > 4$,$n(n-1)(n-2) \neq 0$. Dividing both sides by $n(n-1)(n-2)$,we get:
$(n-3)(n-4) = 42$.
$n^{2} - 7n + 12 = 42$.
$n^{2} - 7n - 30 = 0$.
Factoring the quadratic equation:
$(n-10)(n+3) = 0$.
This gives $n = 10$ or $n = -3$.
Since $n$ must be a positive integer and $n > 4$,we discard $n = -3$.
Therefore,$n = 10$.

(A) Given the equation $\frac{^{n}P_{4}}{^{n-1}P_{4}} = \frac{5}{3}$.
Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:
$\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{(n-5)!}} = \frac{5}{3}$.
This simplifies to $\frac{n \times (n-1)!}{(n-4) \times (n-5)!} \times \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$.
$\frac{n}{n-4} = \frac{5}{3}$.
Cross-multiplying gives $3n = 5(n-4)$.
$3n = 5n - 20$.
$2n = 20$.
$n = 10$.

(A) Given the equation: $5 \times ^{4}P_{r} = 6 \times ^{5}P_{r-1}$
Using the formula $^{n}P_{r} = \frac{n!}{(n-r)!}$,we have:
$5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5!}{(5-(r-1))!} = 6 \times \frac{5!}{(6-r)!}$
Since $5! = 5 \times 4!$,we can write:
$5 \times \frac{4!}{(4-r)!} = 6 \times \frac{5 \times 4!}{(6-r)(5-r)(4-r)!}$
Dividing both sides by $5 \times 4!$:
$\frac{1}{(4-r)!} = \frac{6}{(6-r)(5-r)(4-r)!}$
$1 = \frac{6}{(6-r)(5-r)}$
$(6-r)(5-r) = 6$
$30 - 6r - 5r + r^{2} = 6$
$r^{2} - 11r + 24 = 0$
$(r-8)(r-3) = 0$
Thus,$r = 8$ or $r = 3$.
However,for $^{4}P_{r}$,we must have $r \leq 4$.
Therefore,$r = 8$ is rejected.
Hence,$r = 3$.

(A) The word $DAUGHTER$ consists of $8$ distinct letters: $D, A, U, G, H, T, E, R$.
There are $3$ vowels: $A, U, E$ and $5$ consonants: $D, G, H, T, R$.
Since the vowels must occur together,we treat them as a single unit or object: $(AUE)$.
Now,we have $5$ consonants plus $1$ unit of vowels,making a total of $6$ objects.
These $6$ objects can be arranged in $6!$ ways.
Within the vowel unit,the $3$ vowels $(A, U, E)$ can be arranged among themselves in $3!$ ways.
By the multiplication principle,the total number of arrangements is $6! \times 3!$.
$6! = 720$ and $3! = 6$.
Total arrangements $= 720 \times 6 = 4320$.

(A) The word $DAUGHTER$ contains $8$ distinct letters: $D, A, U, G, H, T, E, R$. The vowels are $A, U, E$ ($3$ vowels) and the consonants are $D, G, H, T, R$ ($5$ consonants).
Total arrangements of $8$ letters $= 8! = 40320$.
To find the arrangements where all vowels are together,treat $(A, U, E)$ as a single unit. Now we have $5$ consonants $+ 1$ unit $= 6$ items,which can be arranged in $6!$ ways. The $3$ vowels within the unit can be arranged in $3!$ ways.
Number of arrangements where vowels are together $= 6! \times 3! = 720 \times 6 = 4320$.
Number of arrangements where vowels do not occur together $=$ Total arrangements $-$ Arrangements where vowels are together.
$= 40320 - 4320 = 36000$.

(A) The total number of discs is $4 + 3 + 2 = 9$.
Since the discs of the same colour are indistinguishable,we use the formula for permutations of a multiset: $\frac{n!}{n_1! n_2! n_3!}$.
Here,$n = 9$,$n_1 = 4$ (red),$n_2 = 3$ (yellow),and $n_3 = 2$ (green).
Therefore,the number of arrangements is $\frac{9!}{4! 3! 2!} = \frac{362880}{24 \times 6 \times 2} = \frac{362880}{288} = 1260$.

(A) The word $INDEPENDENCE$ contains $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.
To find the number of arrangements starting with $P$,we fix $P$ at the first position.
Now,we need to arrange the remaining $11$ letters: $I(1), N(3), D(2), E(4), C(1)$.
The number of arrangements is given by the formula $\frac{n!}{n_1! n_2! n_3! ...}$.
Here,$n = 11$,$n_1 = 3$ (for $N$),$n_2 = 2$ (for $D$),and $n_3 = 4$ (for $E$).
Number of arrangements $= \frac{11!}{3! 2! 4!} = \frac{39916800}{6 \times 2 \times 24} = \frac{39916800}{288} = 138600$.

(A) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.
There are $5$ vowels: $E, E, E, E, I$.
Treating the $5$ vowels as a single object,we have $7$ consonants $(N, N, N, D, D, P, C)$ plus $1$ vowel-block,totaling $8$ objects.
The number of ways to arrange these $8$ objects,where $N$ repeats $3$ times and $D$ repeats $2$ times,is $\frac{8!}{3!2!}$.
The $5$ vowels $(E, E, E, E, I)$ can be arranged among themselves in $\frac{5!}{4!}$ ways.
By the multiplication principle,the total number of arrangements is $\frac{8!}{3!2!} \times \frac{5!}{4!} = \frac{40320}{6 \times 2} \times 5 = 3360 \times 5 = 16800$.

(B) The word $INDEPENDENCE$ has $12$ letters: $I(1), N(3), D(2), E(4), P(1), C(1)$.
Total arrangements $= \frac{12!}{3! \times 2! \times 4!} = \frac{479001600}{6 \times 2 \times 24} = 1663200$.
Treating all vowels $(I, E, E, E, E)$ as one unit,we have $8$ consonants $(N, N, N, D, D, P, C)$ and $1$ vowel unit. Total units $= 9$.
Arrangements with vowels together $= \frac{9!}{3! \times 2!} \times \frac{5!}{4!} = \frac{362880}{6 \times 2} \times 5 = 30240 \times 5 = 151200$.
Number of arrangements where vowels never occur together $= 1663200 - 151200 = 1512000$.

(A) The word $INDEPENDENCE$ has $12$ letters: $I(2), N(3), D(2), E(4), P(1)$.
To find the number of arrangements starting with $I$ and ending with $P$,we fix $I$ at the first position and $P$ at the last position.
Remaining letters are $10$: $N(3), D(2), E(4), I(1), N(0)$ (Wait,let us re-count: $I$ is used once,$P$ is used once).
Remaining letters: $I(1), N(3), D(2), E(4)$.
Total remaining letters = $1 + 3 + 2 + 4 = 10$.
The number of arrangements is given by the formula $\frac{n!}{n_1! n_2! n_3! n_4!}$.
Number of arrangements $= \frac{10!}{1! 3! 2! 4!} = \frac{3628800}{1 \times 6 \times 2 \times 24} = \frac{3628800}{288} = 12600$.

(A) To form a $3$-digit number using the digits $1$ to $9$ without repetition,we need to arrange $3$ distinct digits out of $9$.
The order of the digits matters,so we use the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 9$ and $r = 3$.
Required number of $3$-digit numbers $= ^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!}$.
$= \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = 504$.

(A) $4$-digit number consists of four places: thousands,hundreds,tens,and units.
The thousands place can be filled by any digit from ${1, 2, 3, 4, 5, 6, 7, 8, 9}$. Thus,there are $9$ choices for the thousands place.
The hundreds place can be filled by any of the remaining $9$ digits (including $0$,but excluding the digit used in the thousands place). Thus,there are $9$ choices for the hundreds place.
The tens place can be filled by any of the remaining $8$ digits. Thus,there are $8$ choices for the tens place.
The units place can be filled by any of the remaining $7$ digits. Thus,there are $7$ choices for the units place.
By the multiplication principle,the total number of $4$-digit numbers with no digit repeated is $9 \times 9 \times 8 \times 7 = 4536$.

(C) To form a $3$-digit even number,the units place must be filled with an even digit from the set $\{1, 2, 3, 4, 6, 7\}$.
The available even digits are $2, 4,$ and $6$. Thus,the units place can be filled in $3$ ways.
Since digits cannot be repeated,after filling the units place,we have $5$ digits remaining to fill the hundreds and tens places.
The number of ways to fill the hundreds and tens places using the remaining $5$ digits is given by the permutation formula $^5P_2$.
$^5P_2 = \frac{5!}{(5-2)!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$.
By the multiplication principle,the total number of $3$-digit even numbers is $3 \times 20 = 60$.