Permutation and Combination Questions in English

(B) Total number of ways to arrange $6$ examination papers is $6! = 720$.
If the best and the worst papers appear together,we treat them as a single unit. Thus,we have $5$ units to arrange,which can be done in $5!$ ways. Within the unit,the $2$ papers can be arranged in $2!$ ways.
Number of ways where they appear together = $5! \times 2! = 120 \times 2 = 240$.
Number of ways where they never appear together = (Total arrangements) - (Arrangements where they appear together) = $720 - 240 = 480$.

(C) To form a number between $3000$ and $4000$,the thousands place must be occupied by the digit $3$. This is $1$ way.
For the number to be divisible by $5$,the units place must be occupied by the digit $5$. This is $1$ way.
The remaining two places (hundreds and tens) must be filled by the remaining digits from the set ${1, 2, 4, 6}$.
There are $4$ remaining digits and $2$ places to fill.
The number of ways to arrange these is given by $^4P_2 = 4 \times 3 = 12$.
Thus,the total number of such numbers is $12$,which corresponds to $^4P_2$.

(A) Each of the $6$ distinct rings can be worn on any of the $4$ fingers.
Since each ring has $4$ independent choices of fingers,the total number of ways to wear the $6$ rings is $4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^6$.
Therefore,the correct option is $A$.

(D) To form numbers using the digits $1, 2, 3, 4$ without repetition,we can form numbers of different lengths:
- Number of $1$-digit numbers $= ^4P_1 = 4$
- Number of $2$-digit numbers $= ^4P_2 = 4 \times 3 = 12$
- Number of $3$-digit numbers $= ^4P_3 = 4 \times 3 \times 2 = 24$
- Number of $4$-digit numbers $= ^4P_4 = 4 \times 3 \times 2 \times 1 = 24$
Since these cases are mutually exclusive,the total number of possible numbers is the sum of these values:
Total $= ^4P_1 + ^4P_2 + ^4P_3 + ^4P_4$.

(B) Each of the $7$ men can cast their vote for any one of the $3$ candidates.
Since each man has $3$ independent choices,the $1^{st}$ man has $3$ ways to vote,the $2^{nd}$ man has $3$ ways,and so on,up to the $7^{th}$ man.
Therefore,the total number of ways the votes can be given is $3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^7$ ways.

(A) The man has $4$ options to travel from Gwalior to Bhopal.
Since he must return by a different bus,he has $4 - 1 = 3$ options for the return journey.
According to the fundamental principle of counting,the total number of ways is the product of the number of choices for each step.
Total ways = $4 \times 3 = 12$.

(B) Given the equation: ${}^n{P_5} = 20 \cdot {}^n{P_3}$
Using the formula ${}^n{P_r} = \frac{n!}{(n-r)!}$,we have:
$\frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}$
Dividing both sides by $n!$ (since $n! \neq 0$):
$\frac{1}{(n-5)!} = \frac{20}{(n-3)!}$
$(n-3)! = 20 \cdot (n-5)!$
$(n-3)(n-4)(n-5)! = 20 \cdot (n-5)!$
Since $(n-5)! \neq 0$,we can cancel it from both sides:
$(n-3)(n-4) = 20$
$n^2 - 7n + 12 = 20$
$n^2 - 7n - 8 = 0$
$(n-8)(n+1) = 0$
This gives $n = 8$ or $n = -1$.
Since $n$ must be a positive integer and $n \ge 5$ for ${}^n{P_5}$ to be defined,we discard $n = -1$.
Therefore,$n = 8$.

(A) The word $UNIVERSAL$ contains $9$ distinct letters: $U, N, I, V, E, R, S, A, L$.
To form a word of $3$ letters from these $9$ distinct letters,we use the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 9$ and $r = 3$.
Therefore,the number of words = $^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 9 \times 8 \times 7 = 504$.

(C) The formula for permutation is ${}^n{P_r} = \frac{n!}{(n-r)!}$.
Given the ratio: $\frac{{}^n{P_4}}{{}^n{P_5}} = \frac{1}{2}$.
Substituting the formula: $\frac{n!}{(n-4)!} \div \frac{n!}{(n-5)!} = \frac{1}{2}$.
This simplifies to: $\frac{n!}{(n-4)!} \times \frac{(n-5)!}{n!} = \frac{1}{2}$.
Canceling $n!$ from numerator and denominator: $\frac{(n-5)!}{(n-4)!} = \frac{1}{2}$.
Since $(n-4)! = (n-4) \times (n-5)!$,we have: $\frac{(n-5)!}{(n-4)(n-5)!} = \frac{1}{2}$.
$\frac{1}{n-4} = \frac{1}{2}$.
Therefore,$n-4 = 2$,which gives $n = 6$.

(C) Each of the $mn$ letters can be posted in any of the $n$ letter-boxes.
Since each letter has $n$ independent choices,the total number of ways to post $mn$ letters is $n \times n \times n \times \dots \times n$ ($mn$ times).
Therefore,the total number of ways is $n^{mn}$.

(D) Each true-false question has $2$ possible outcomes: either 'True' or 'False'.
Since there are $10$ independent questions,the total number of ways to answer them is calculated by multiplying the number of choices for each question.
Total ways = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}$.
Calculating $2^{10}$,we get $1024$.
Therefore,the correct option is $D$.

(A) number is even if its unit digit is even. The available digits are ${1, 2, 3, 4, 5, 6, 7, 8, 9}$.
The even digits available are ${2, 4, 6, 8}$.
Step $1$: The unit digit can be filled in $4$ ways (choosing from $2, 4, 6, 8$).
Step $2$: Since repetition is not allowed,we have $8$ remaining digits to fill the remaining $2$ places.
Step $3$: The number of ways to arrange $2$ digits out of the remaining $8$ is given by $^8P_2 = 8 \times 7 = 56$.
Step $4$: Total even numbers $= 4 \times 56 = 224$.

(D) Given the equation: $^nP_5 = 9 \times ^{n-1}P_4$
Using the formula for permutations,$^nP_r = \frac{n!}{(n-r)!}$,we have:
$\frac{n!}{(n-5)!} = 9 \times \frac{(n-1)!}{((n-1)-4)!}$
$\frac{n \times (n-1)!}{(n-5)!} = 9 \times \frac{(n-1)!}{(n-5)!}$
Since $(n-1)!$ and $(n-5)!$ are non-zero,we can cancel them from both sides:
$n = 9$
Thus,the value of $n$ is $9$.

(A) The correct option is $A$.
We know that $^n{P_r} = \frac{n!}{(n-r)!}$.
Consider the expression in option $A$: $^{n-1}{P_r} + r \cdot {^{n-1}}{P_{r-1}}$.
$= \frac{(n-1)!}{(n-1-r)!} + r \cdot \frac{(n-1)!}{(n-1-(r-1))!}$
$= \frac{(n-1)!}{(n-r-1)!} + r \cdot \frac{(n-1)!}{(n-r)!}$
$= \frac{(n-1)!}{(n-r-1)!} \left[ 1 + \frac{r}{n-r} \right]$
$= \frac{(n-1)!}{(n-r-1)!} \left[ \frac{n-r+r}{n-r} \right]$
$= \frac{(n-1)!}{(n-r-1)!} \cdot \frac{n}{n-r}$
$= \frac{n \cdot (n-1)!}{(n-r) \cdot (n-r-1)!} = \frac{n!}{(n-r)!} = ^n{P_r}$.

(A) There are $10$ digits in total,namely $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
To form a $9$ digit number with all distinct digits,the first digit cannot be $0$.
For the first position,there are $9$ choices ($1$ to $9$).
For the remaining $8$ positions,we need to choose $8$ digits from the remaining $9$ digits (including $0$),which can be arranged in $P(9, 8)$ ways.
Total numbers = $9 \times P(9, 8) = 9 \times \frac{9!}{(9-8)!} = 9 \times 9!$.
Alternatively,total $9$ digit numbers with distinct digits = (Total permutations of $9$ digits out of $10$) - (Permutations starting with $0$):
$= P(10, 9) - P(9, 8) = \frac{10!}{1!} - \frac{9!}{1!} = 10! - 9! = (10 - 1) \times 9! = 9 \times 9!$.

(C) The total number of outcomes when $4$ dice are rolled is $6^4 = 1296$.
To find the number of outcomes where at least one die shows $2$,we subtract the number of outcomes where $2$ does not appear on any die from the total number of outcomes.
If $2$ does not appear on any die,each die can show any of the remaining $5$ faces $(1, 3, 4, 5, 6)$.
Thus,the number of outcomes where $2$ does not appear is $5^4 = 625$.
Therefore,the number of outcomes with at least one $2$ is $1296 - 625 = 671$.

(C) Each of the $4$ parcels can be sent to any of the $5$ post-offices.
Since each parcel is independent and has $5$ choices for registration,the total number of ways is calculated by multiplying the number of choices for each parcel.
Total ways = $5 \times 5 \times 5 \times 5 = 5^4 = 625$.

(A) Each of the $5$ distinct prizes can be given to any of the $4$ students.
Since each prize has $4$ independent choices,the total number of ways to distribute the $5$ prizes is $4 \times 4 \times 4 \times 4 \times 4 = 4^5$.
Calculating the value,$4^5 = 1024$.
Therefore,the total number of ways is $1024$.

(D) To find the number of ways $3$ passengers can sit in $5$ vacant seats,we use the concept of permutations because the order in which they sit matters.
The number of ways to arrange $r$ objects out of $n$ distinct objects is given by the formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 5$ (total seats) and $r = 3$ (passengers).
Therefore,the number of ways is $^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$.
Thus,there are $60$ ways for the passengers to sit.

(A) The product of $r$ consecutive natural numbers is given by $n(n+1)(n+2)...(n+r-1)$.
This expression is equal to $r! \times \binom{n+r-1}{r}$.
Since the binomial coefficient $\binom{n+r-1}{r}$ is always an integer,the product of $r$ consecutive natural numbers must be divisible by $r!$.

(C) To find the sum of the digits in the unit place of all numbers formed using the digits $3, 4, 5, 6$ taken all at a time:
$1$. The total number of permutations of $4$ distinct digits taken all at a time is $4! = 24$.
$2$. If we fix one digit (e.g.,$3$) in the unit place,the remaining $3$ digits can be arranged in the other $3$ positions in $3! = 6$ ways.
$3$. This means each digit $(3, 4, 5, 6)$ appears in the unit place exactly $6$ times.
$4$. The sum of the digits in the unit place is $6 \times (3 + 4 + 5 + 6) = 6 \times 18 = 108$.

(A) The total number of coins is $6$.
For the number of tails to be equal to the number of heads,we must have $3$ heads and $3$ tails.
Since the coins are identical,the number of ways to arrange $3$ heads and $3$ tails in a row is given by the formula for permutations of a multiset: $\frac{n!}{n_1! n_2!} = \frac{6!}{3! 3!}$.
Calculating this: $\frac{720}{6 \times 6} = \frac{720}{36} = 20$.
Thus,there are $20$ ways.

(C) Total number of $5$-digit numbers formed using the digits ${4, 5, 6, 7, 8}$ without repetition is $5! = 120$.
To find the numbers greater than $56000$,we subtract the numbers less than or equal to $56000$ from the total.
Numbers starting with $4$: There are $4! = 24$ such numbers.
Numbers starting with $54$: There are $3! = 6$ such numbers.
Numbers starting with $55$: Not possible as digits are distinct.
Numbers starting with $56000$: Since the digits are ${4, 5, 6, 7, 8}$,the smallest number starting with $56$ is $56478$,which is already greater than $56000$.
Thus,the numbers less than $56000$ are those starting with $4$ ($24$ numbers) and those starting with $54$ ($6$ numbers).
Total numbers less than $56000 = 24 + 6 = 30$.
Required numbers greater than $56000 = 120 - 30 = 90$.

(C) Let the two boys be $A$ and $B$.
Case $1$: Boy $A$ receives $2$ balls and boy $B$ receives $8$ balls. The number of ways to choose $2$ balls out of $10$ for boy $A$ is given by $\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$.
Case $2$: Boy $A$ receives $8$ balls and boy $B$ receives $2$ balls. The number of ways to choose $8$ balls out of $10$ for boy $A$ is given by $\binom{10}{8} = \frac{10!}{8!2!} = 45$.
Since these are two distinct scenarios,the total number of ways is $45 + 45 = 90$.

(A) The total number of $4$ digit numbers that can be formed using $4$ distinct digits is $4! = 24$.
Each digit appears in each place (units,tens,hundreds,thousands) an equal number of times,which is $24 / 4 = 6$ times.
The sum of the given digits is $2 + 4 + 6 + 8 = 20$.
The sum of the digits in any given place is $6 \times 20 = 120$.
Therefore,the sum of all such numbers is $120 \times 1 + 120 \times 10 + 120 \times 100 + 120 \times 1000$.
$= 120(1 + 10 + 100 + 1000) = 120 \times 1111 = 133320$.

(A) The villager has $5$ choices to go from the village to the town.
Since the villager can return by any of the $5$ roads,there are also $5$ choices for the return journey.
According to the fundamental principle of counting,the total number of ways to perform both tasks (going and returning) is the product of the number of ways to perform each task.
Total ways = $5 \times 5 = 25$.

(D) Total number of arrangements of $5$ distinct examination papers is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
To find the number of ways where physics and chemistry papers never come together,we first calculate the number of ways they $DO$ come together.
Treat the physics and chemistry papers as a single unit. Now,we have $4$ units to arrange (the single unit of physics-chemistry + the remaining $3$ papers),which can be done in $4!$ ways.
Within the single unit,the physics and chemistry papers can be arranged in $2!$ ways.
Therefore,the number of arrangements where they come together is $4! \times 2! = 24 \times 2 = 48$.
The number of ways they never come together is the total arrangements minus the arrangements where they are together: $120 - 48 = 72$.

(B) The first prize can be awarded to any of the $5$ competitors in $5$ ways.
After the first prize is awarded,the second prize can be awarded to any of the remaining $4$ competitors in $4$ ways.
Finally,the third prize can be awarded to any of the remaining $3$ competitors in $3$ ways.
Since a competitor cannot receive more than one prize,the total number of ways is calculated by the product of these choices.
Total ways $= 5 \times 4 \times 3 = 60$ ways.

(B) $3$-digit number has three places: hundreds,tens,and units.
For the number to be odd,the unit place must be filled by an odd digit from the given set ${1, 2, 3, 4, 5, 6}$. The odd digits available are ${1, 3, 5}$. Thus,the unit place can be filled in $3$ ways.
Since repetition is allowed,the hundreds place can be filled by any of the $6$ digits ${1, 2, 3, 4, 5, 6}$,which gives $6$ ways.
The tens place can also be filled by any of the $6$ digits ${1, 2, 3, 4, 5, 6}$,which gives $6$ ways.
Using the fundamental principle of counting,the total number of such odd numbers $= 6 \times 6 \times 3 = 108$.

(A) The given digits are $2, 0, 4, 3, 8$. There are $5$ distinct digits.
$A$ five-digit number cannot have $0$ in the ten-thousands place.
Total permutations of $5$ digits taken all at a time is $5! = 120$.
Permutations where $0$ is in the ten-thousands place (fixing $0$ at the first position,we arrange the remaining $4$ digits in the remaining $4$ places) is $4! = 24$.
Therefore,the number of five-digit numbers that can be formed is $5! - 4! = 120 - 24 = 96$.

(C) The formula for permutations is given by $^{n}P_r = \frac{n!}{(n-r)!}$.
Given $^{12}P_r = 1320$.
We can write this as $12 \times 11 \times 10 = 1320$.
Since $12 \times 11 \times 10$ represents the product of $3$ consecutive integers starting from $12$,it is equivalent to $^{12}P_3$.
Therefore,by comparing $^{12}P_r$ with $^{12}P_3$,we get $r = 3$.

(A) For an $n$-digit number,the first digit (at the leftmost position) can be any digit from $1$ to $9$ (since it cannot be $0$). Thus,there are $9$ choices for the first digit.
For the second digit,it can be any digit from $0$ to $9$ except the one used for the first digit. Thus,there are $9$ choices for the second digit.
For the third digit,it can be any digit from $0$ to $9$ except the one used for the second digit. Thus,there are $9$ choices for the third digit.
Continuing this pattern,for each subsequent position up to the $n$-th digit,there are always $9$ choices available because each digit only needs to be different from the one immediately preceding it.
Therefore,the total number of $n$-digit numbers is $9 \times 9 \times 9 \times \dots \times 9$ ($n$ times),which equals $9 \times 9^{n-1}$.

(C) The word $SALOON$ contains $6$ letters,where $O$ repeats $2$ times.
Total number of arrangements of the letters of the word $SALOON$ is given by $\frac{6!}{2!} = \frac{720}{2} = 360$.
To find the number of arrangements where the two $O$'s come together,we treat the two $O$'s as a single unit $(OO)$. Now,we have $5$ units: $S, A, L, N, (OO)$.
The number of arrangements of these $5$ units is $5! = 120$.
Therefore,the number of arrangements where the two $O$'s do not come together is the total arrangements minus the arrangements where they are together:
$360 - 120 = 240$.

(A) The word $MAXIMUM$ contains $7$ letters: $M, A, X, I, M, U, M$.
Consonants are $M, X, M, M$ ($4$ letters) and vowels are $A, I, U$ ($3$ letters).
To ensure no two consonants occur together,we use the gap method.
First,arrange the $3$ vowels $(A, I, U)$ in $3!$ ways.
These $3$ vowels create $4$ possible gaps (including ends): $\_ V \_ V \_ V \_$.
We need to place the $4$ consonants $(M, X, M, M)$ into these $4$ gaps.
The number of ways to arrange $4$ consonants in $4$ gaps is $\frac{4!}{3!}$ (since $M$ repeats $3$ times).
Total number of words = (Arrangement of vowels) $\times$ (Arrangement of consonants in gaps)
$= 3! \times \frac{4!}{3!} = 4! = 24$.

(B) The total number of ways to arrange $n$ books in a row is $n!$.
If two specified books are always together,we can treat them as a single unit. This leaves us with $(n - 1)$ units to arrange,which can be done in $(n - 1)!$ ways.
Within the single unit,the two specified books can be arranged in $2! = 2$ ways.
Therefore,the number of ways the two specified books are always together is $2 \times (n - 1)!$.
To find the number of ways where the two specified books are not together,we subtract the number of ways they are together from the total number of arrangements:
Required ways $= n! - 2(n - 1)!$
$= n \times (n - 1)! - 2(n - 1)!$
$= (n - 1)! (n - 2)$.

(A) To form a number between $500$ and $600$,the digit at the hundreds place must be $5$.
Since the digits cannot be repeated,we have fixed the hundreds place with the digit $5$.
The remaining digits available are ${1, 2, 3, 4, 6}$,which gives us $5$ choices.
We need to fill the tens place and the units place using these $5$ remaining digits.
The number of ways to arrange $2$ digits out of $5$ is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 5$ and $r = 2$,so $^5P_2 = \frac{5!}{(5-2)!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$.
Therefore,there are $20$ such numbers.

(B) The numbers must be $4$-digit numbers since they are between $1000$ and $4000$.
Let the $4$-digit number be represented as $d_1 d_2 d_3 d_4$.
For the number to be greater than $1000$ and less than or equal to $4000$,the first digit $d_1$ can be $1, 2, 3,$ or $4$.
Case $1$: If $d_1 = 1, 2,$ or $3$,then each of the remaining three positions $(d_2, d_3, d_4)$ can be filled by any of the $5$ digits $(0, 1, 2, 3, 4)$ in $5$ ways each.
Number of such integers = $3 \times 5 \times 5 \times 5 = 375$.
However,this includes the number $1000$ (where $d_1=1, d_2=0, d_3=0, d_4=0$). Since the question asks for numbers strictly greater than $1000$,we must exclude $1000$.
So,count = $375 - 1 = 374$.
Case $2$: If $d_1 = 4$,the only number possible is $4000$ because any other combination like $4001$ would be greater than $4000$.
Thus,we add $1$ for the number $4000$.
Total numbers = $374 + 1 = 375$.

(B) The given digits are $1, 2, 3, 4, 3, 2, 1$. There are $7$ digits in total.
The odd digits are $1, 3, 3, 1$ (total $4$ digits).
The even digits are $2, 4, 2$ (total $3$ digits).
The odd places are $1^{st}, 3^{rd}, 5^{th}, 7^{th}$ (total $4$ places).
The even places are $2^{nd}, 4^{th}, 6^{th}$ (total $3$ places).
Since odd digits must occupy odd places,we arrange $1, 3, 3, 1$ in $4$ odd places. The number of ways is $\frac{4!}{2!2!} = \frac{24}{4} = 6$.
Next,we arrange the even digits $2, 4, 2$ in $3$ even places. The number of ways is $\frac{3!}{2!} = 3$.
Therefore,the total number of ways is $6 \times 3 = 18$.

(C) Step $1$: Arrange the $5$ boys in a row. The number of ways to arrange $5$ boys is $5!$.
Step $2$: To ensure no two girls sit together,we place the girls in the gaps created by the boys. With $5$ boys,there are $6$ possible gaps (including the ends) where the $3$ girls can be seated.
Step $3$: The number of ways to choose and arrange $3$ girls in these $6$ gaps is given by the permutation formula $^6P_3$.
Step $4$: The total number of ways is the product of the arrangements of boys and the arrangements of girls: $5! \times ^6P_3$.

(A) Numbers less than $1000$ can be $1$-digit,$2$-digit,or $3$-digit numbers.
$1$-digit numbers: We can choose any of the $6$ digits. Number of ways $= ^6P_1 = 6$.
$2$-digit numbers: We can choose $2$ digits out of $6$ where order matters. Number of ways $= ^6P_2 = 6 \times 5 = 30$.
$3$-digit numbers: We can choose $3$ digits out of $6$ where order matters. Number of ways $= ^6P_3 = 6 \times 5 \times 4 = 120$.
Total numbers $= 6 + 30 + 120 = 156$.

(A) The word $COURTESY$ consists of $8$ distinct letters: $C, O, U, R, T, E, S, Y$.
We need to form words where the first letter is $C$ and the last letter is $Y$.
Fixing $C$ at the first position and $Y$ at the last position leaves $8 - 2 = 6$ letters to be arranged in the middle positions.
The number of ways to arrange these $6$ remaining letters is $6!$.
Therefore,the total number of words is $6!$.

(B) The word $DELHI$ consists of $5$ distinct letters: $D, E, L, H, I$.
Since the letter $L$ must be fixed in the middle position,we have $1$ position filled out of $5$.
The remaining $4$ positions can be filled by the remaining $4$ letters $(D, E, H, I)$ in $4!$ ways.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,the total number of words that can be formed is $24$.

(B) To form a $5$-digit number using the digits $3, 4, 7$ (each once) and $5$ (twice),we need to arrange the set of digits ${3, 4, 7, 5, 5}$.
The total number of arrangements of $n$ objects where $n_1$ objects are of one kind,$n_2$ are of another kind,etc.,is given by the formula $\frac{n!}{n_1! n_2! \dots}$.
Here,the total number of digits $n = 5$.
The digit $5$ repeats $2$ times,so $n_1 = 2$.
The digits $3, 4,$ and $7$ appear $1$ time each.
Therefore,the number of ways is $\frac{5!}{2!} = \frac{120}{2} = 60$.

(B) Each of the $3$ letters can be posted in any of the $4$ letter-boxes.
Since each letter has $4$ choices,the total number of ways to post $3$ letters in $4$ letter-boxes is $4 \times 4 \times 4 = 4^3 = 64$.
However,the condition states that all letters should not be posted in the same letter-box.
There are $4$ cases where all $3$ letters are posted in the same box (i.e.,all in box $1$,all in box $2$,all in box $3$,or all in box $4$).
Therefore,the required number of ways is $64 - 4 = 60$.

(D) The total number of $5$ digit telephone numbers that can be formed using the digits $0, 1, 2, ..., 9$ is $10^5 = 100000$ (since each of the $5$ positions can be filled by any of the $10$ digits).
The number of $5$ digit telephone numbers where no digit is repeated is given by the permutation formula $^{10}P_5 = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
The number of telephone numbers having at least one digit repeated is the total number of telephone numbers minus the number of telephone numbers with no repeated digits.
Required number $= 100000 - 30240 = 69760$.

(C) 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 formula for the number of permutations of $n$ objects where $n_1, n_2, \dots, n_k$ are the frequencies of identical objects is given by $\frac{n!}{n_1! n_2! \dots n_k!}$.
Substituting the values: $n = 11$,$n_1 = 2$ (for $M$),$n_2 = 2$ (for $A$),and $n_3 = 2$ (for $T$).
Required number of words = $\frac{11!}{2! 2! 2!}$.

(B) The word $CALCUTTA$ contains $8$ letters in total.
The frequency of each letter is as follows:
$C$ appears $2$ times.
$A$ appears $2$ times.
$L$ appears $1$ time.
$U$ appears $2$ times.
$T$ appears $2$ times.
Wait,let us re-count: $C(2), A(2), L(1), U(2), T(2)$. Total letters = $2+2+1+2+2 = 9$.
Correction: The word $CALCUTTA$ has $8$ letters: $C, A, L, C, U, T, T, A$.
Frequency: $C=2, A=2, L=1, U=1, T=2$.
Total letters $n = 8$.
Number of arrangements = $\frac{n!}{n_1! n_2! ... n_k!} = \frac{8!}{2! 2! 2!} = \frac{40320}{2 \times 2 \times 2} = \frac{40320}{8} = 5040$.

(A) The numbers lying between $99$ and $1000$ are all $3$-digit numbers.
We have $6$ digits available: ${0, 2, 3, 6, 7, 8}$.
$A$ $3$-digit number cannot have $0$ at the hundreds place.
Total ways to arrange $3$ digits out of $6$ is given by $^6P_3 = 6 \times 5 \times 4 = 120$.
However,we must exclude the cases where $0$ is at the hundreds place. If $0$ is fixed at the hundreds place,we need to arrange $2$ more digits from the remaining $5$ digits in the tens and units places. This is given by $^5P_2 = 5 \times 4 = 20$.
Therefore,the total number of valid $3$-digit numbers is $120 - 20 = 100$.

(B) Let the $10$ animals be $A_1, A_2, \dots, A_{10}$ and the $10$ cages be $C_1, C_2, \dots, C_{10}$.
Let $C_1, C_2, C_3, C_4$ be the $4$ small cages.
Let $A_1, A_2, A_3, A_4, A_5$ be the $5$ animals that cannot enter the small cages.
These $5$ animals must be placed in the remaining $6$ cages $(C_5, C_6, C_7, C_8, C_9, C_{10})$.
The number of ways to arrange these $5$ animals in $6$ cages is given by the permutation formula $^6P_5 = \frac{6!}{(6-5)!} = 6! = 720$.
After placing these $5$ animals, we have $5$ animals remaining and $5$ cages remaining (the $4$ small cages plus the $1$ cage left over from the $6$ large cages).
The number of ways to arrange the remaining $5$ animals in the remaining $5$ cages is $5! = 120$.
Therefore, the total number of ways = $^6P_5 \times 5! = 720 \times 120 = 86400$.

(B) The word $COMMITTEE$ consists of $9$ letters in total.
The frequency of each letter is as follows:
$C: 2$
$O: 1$
$M: 2$
$I: 1$
$T: 2$
$E: 1$
Total number of arrangements = $\frac{n!}{n_1! n_2! n_3! ...}$
Here,$n = 9$,$n_1 = 2$ (for $C$),$n_2 = 2$ (for $M$),and $n_3 = 2$ (for $T$).
Therefore,the total number of words = $\frac{9!}{2! \times 2! \times 2!} = \frac{9!}{(2!)^3}$.