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

(C) When we arrange $1$ thing at a time,the number of possible permutations is $n$.
When we arrange them $2$ at a time,the number of possible permutations is $n \times n = n^2$.
Continuing this process up to $r$ at a time,the number of permutations for $k$ things is $n^k$.
Thus,the total number of permutations taken not more than $r$ at a time is the sum of a geometric progression:
$n + n^2 + \ldots + n^r = \frac{n(n^r - 1)}{n - 1}$ (where $n > 1$).

(D) We know that ${}^n P_r = \frac{n!}{(n-r)!}$.
Consider the expression ${}^n P_r - {}^{n-1} P_r = \frac{n!}{(n-r)!} - \frac{(n-1)!}{(n-r-1)!}$.
$= \frac{(n-1)!}{(n-r-1)!} \left( \frac{n}{n-r} - 1 \right) = \frac{(n-1)!}{(n-r-1)!} \left( \frac{n - (n-r)}{n-r} \right)$.
$= \frac{(n-1)!}{(n-r-1)!} \cdot \frac{r}{n-r} = \frac{r \cdot (n-1)!}{(n-r)!}$.
$= r \cdot \frac{(n-1)!}{((n-1) - (r-1))!} = r \cdot {}^{n-1} P_{r-1}$.
Thus,${}^n P_r = {}^{n-1} P_r + r \cdot {}^{n-1} P_{r-1}$.
Comparing this with the given equation ${}^n P_r = {}^{n-1} P_r + x \cdot {}^{n-1} P_{r-1}$,we get $x = r$.

(B) Given expression: $\frac{1}{r+1}\left\{{ }^n P_{r+1}-{ }^{(n-1)} P_{r+1}\right\}$
Using the formula ${ }^n P_r = \frac{n!}{(n-r)!}$,we have:
$= \frac{1}{r+1} \left[ \frac{n!}{(n-(r+1))!} - \frac{(n-1)!}{(n-1-(r+1))!} \right]$
$= \frac{1}{r+1} \left[ \frac{n!}{(n-r-1)!} - \frac{(n-1)!}{(n-r-2)!} \right]$
$= \frac{1}{r+1} \left[ \frac{n(n-1)!}{(n-r-1)!} - \frac{(n-1)!(n-r-1)}{(n-r-1)!} \right]$
$= \frac{(n-1)!}{(r+1)(n-r-1)!} [n - (n-r-1)]$
$= \frac{(n-1)!}{(r+1)(n-r-1)!} [r+1]$
$= \frac{(n-1)!}{(n-r-1)!} = { }^{n-1} P_r$

(C) We have,
${}^7P_3 - 3({}^6P_2)$
$= \frac{7!}{(7-3)!} - 3 \times \frac{6!}{(6-2)!}$
$= \frac{7!}{4!} - 3 \times \frac{6!}{4!}$
$= \frac{7 \times 6!}{4!} - \frac{3 \times 6!}{4!}$
$= \frac{6!}{4!} (7 - 3)$
$= \frac{6!}{4!} \times 4$
$= \frac{6 \times 5 \times 4!}{4!} \times 4$
$= 30 \times 4 = 120$
Alternatively,using the definition of permutation:
${}^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120$
Thus,the value is equal to ${}^6P_3$.

(C) The letters in '$GOVIND$' are $D, G, I, N, O, V$. Total letters = $6$. All letters are distinct.
Total number of words = $6! = 720$.
To find the rank of '$GOVIND$',we arrange the letters in alphabetical order: $D, G, I, N, O, V$.
$1$. Words starting with $D$: $5! = 120$.
$2$. Words starting with $G$:
- $GD...$: $4! = 24$.
- $GI...$: $4! = 24$.
- $GN...$: $4! = 24$.
- $GO...$: Next letter is $D$ (alphabetical order).
- $GOD...$: $3! = 6$.
- $GOI...$: $3! = 6$.
- $GON...$: $3! = 6$.
- $GOV...$: Next letter is $D$ (alphabetical order).
- $GOVD...$: $2! = 2$.
- $GOVI...$: Next letter is $D$ (alphabetical order).
- $GOVIDN$: $1$.
- $GOVIND$: $1$.
Rank of '$GOVIND$' = $120 + (24 \times 3) + (6 \times 3) + 2 + 1 + 1 = 120 + 72 + 18 + 4 = 214$.
Number of words after '$GOVIND$' = $720 - 214 = 506$.

(B) The letters of the word $REPEAT$ are $\{A, E, E, P, R, T\}$.
Arranging these letters in alphabetical order,we get: $A, E, E, P, R, T$.
$1$. Words starting with $A$: $\frac{5!}{2!} = 60$.
$2$. Words starting with $E$: $5! = 120$.
$3$. Words starting with $P$: $\frac{5!}{2!} = 60$.
$4$. Words starting with $RA$: $\frac{4!}{2!} = 12$.
$5$. Words starting with $REA$: $3! = 6$.
$6$. Words starting with $REE$: $3! = 6$.
$7$. Words starting with $REPA$: $2! = 2$.
$8$. The next word is $REPEAT$: $1$.
Summing these up: $60 + 120 + 60 + 12 + 6 + 6 + 2 + 1 = 267$.
Thus,the rank of the word $REPEAT$ is $267$.

(C) The word $SARANAM$ contains $7$ letters: $A, A, A, S, R, N, M$. The distinct letters are $\{A, S, R, N, M\}$.
We need to form $5$-letter words. The possible cases are:
$(i)$ All $5$ letters are different (using $S, A, R, N, M$):
Number of words $= 5! = 120$.
(ii) $2$ letters are alike (specifically $A$) and $3$ are different:
We choose $3$ letters from the remaining $4$ distinct letters $\{S, R, N, M\}$ in $^4C_3$ ways. The number of arrangements is $\frac{5!}{2!} = 60$.
Total words $= ^4C_3 \times 60 = 4 \times 60 = 240$.
(iii) $3$ letters are alike (specifically $A$) and $2$ are different:
We choose $2$ letters from the remaining $4$ distinct letters $\{S, R, N, M\}$ in $^4C_2$ ways. The number of arrangements is $\frac{5!}{3!} = 20$.
Total words $= ^4C_2 \times 20 = 6 \times 20 = 120$.
Total number of words $= 120 + 240 + 120 = 480$.

(A) The digits are $\{2, 3, 4, 5, 6\}$. Arranging them in decreasing order: $6, 5, 4, 3, 2$.
Total permutations possible = $5! = 120$.
We want the $89^{th}$ number when arranged in decreasing order.
Numbers starting with $6$: $4! = 24$ numbers (Ranks $1$ to $24$).
Numbers starting with $5$: $4! = 24$ numbers (Ranks $25$ to $48$).
Numbers starting with $4$: $4! = 24$ numbers (Ranks $49$ to $72$).
Numbers starting with $36$: $3! = 6$ numbers (Ranks $73$ to $78$).
Numbers starting with $35$: $3! = 6$ numbers (Ranks $79$ to $84$).
Numbers starting with $346$: $2! = 2$ numbers (Ranks $85$ to $86$).
Numbers starting with $345$: $2! = 2$ numbers (Ranks $87$ to $88$).
The $89^{th}$ number starts with $34265$ (Rank $89$) and the $90^{th}$ is $34256$ (Rank $90$).
Thus,the $89^{th}$ number is $34265$.

(A) number is divisible by $3$ if the sum of its digits is divisible by $3$.
Let the four-digit number be $d_1 d_2 d_3 d_4$.
There are $9$ choices for each of the first three digits $(d_1, d_2, d_3)$,which are ${1, 2, 3, 4, 5, 6, 7, 8, 9}$.
Total ways to choose the first three digits is $9 \times 9 \times 9 = 729$.
Let the sum of the first three digits be $S = d_1 + d_2 + d_3$.
For the whole number to be divisible by $3$,$S + d_4$ must be a multiple of $3$.
For any value of $S$,we check the possible values of $d_4 \in {1, 2, 3, 4, 5, 6, 7, 8, 9}$ such that $S + d_4 \equiv 0 \pmod{3}$.
If $S \equiv 0 \pmod{3}$,then $d_4 \in {3, 6, 9}$ ($3$ choices).
If $S \equiv 1 \pmod{3}$,then $d_4 \in {2, 5, 8}$ ($3$ choices).
If $S \equiv 2 \pmod{3}$,then $d_4 \in {1, 4, 7}$ ($3$ choices).
In all cases,there are exactly $3$ choices for $d_4$.
Therefore,the total number of such four-digit numbers is $729 \times 3 = 2187$.

(C) The word $BANANA$ contains $6$ letters: $A, A, A, B, N, N$.
Total arrangements of $BANANA = \frac{6!}{3!2!1!} = \frac{720}{6 \times 2} = 60$.
To find the number of arrangements where the two $N$s do not come together,we subtract the arrangements where the two $N$s are together from the total arrangements.
Treat the two $N$s as a single unit $(NN)$. Now we have $5$ units: $A, A, A, B, (NN)$.
Arrangements with $N$s together $= \frac{5!}{3!1!1!} = \frac{120}{6} = 20$.
Number of ways where $N$s do not come together $= 60 - 20 = 40$.

(B) five-digit number is divisible by $5$ if its unit place is either $0$ or $5$.
Case $I$: When $0$ is in the unit place,the remaining $4$ positions can be filled by the remaining $5$ digits $(1, 2, 3, 4, 5)$ in $^5P_4 = 120$ ways.
Case $II$: When $5$ is in the unit place,the first position (ten-thousands place) cannot be $0$. Thus,the first position can be filled by any of the $4$ digits $(1, 2, 3, 4)$. The remaining $3$ positions can be filled by the remaining $4$ digits (including $0$) in $^4P_3$ ways.
Number of ways for Case $II = 4 \times ^4P_3 = 4 \times 24 = 96$.
Total number of ways $= 120 + 96 = 216$.

(C) The total number of words of length $4$ that can be formed using $8$ different letters (with repetition allowed) is $8^{4}$.
The number of words of length $4$ that can be formed using $8$ different letters without any repetition is given by the permutation formula ${}^{8}P_{4}$.
The number of words with at least one letter repeated is the total number of words minus the number of words with no letters repeated.
Therefore,the required number of words is $8^{4} - {}^{8}P_{4}$.

(A) Natural numbers less than $1000$ can be $1$-digit,$2$-digit,or $3$-digit numbers.
Case $1$: $1$-digit numbers: The digits can be $1, 2, 3, 4, 5, 6, 7, 8, 9$. Total $= 9$.
Case $2$: $2$-digit numbers: The tens place can be filled in $9$ ways (excluding $0$) and the units place can be filled in $9$ ways (including $0$ but excluding the digit used in the tens place). Total $= 9 \times 9 = 81$.
Case $3$: $3$-digit numbers: The hundreds place can be filled in $9$ ways (excluding $0$),the tens place in $9$ ways (including $0$ but excluding the digit used in the hundreds place),and the units place in $8$ ways (excluding the digits used in the hundreds and tens places). Total $= 9 \times 9 \times 8 = 648$.
Total number of natural numbers $= 9 + 81 + 648 = 738$.

(D) There are $10$ speakers in total. The total number of ways to arrange $10$ speakers is $10!$.
In any arrangement,there are two possibilities for the relative order of $S_1$ and $S_2$: either $S_1$ speaks before $S_2$,or $S_2$ speaks before $S_1$.
Since the condition is that $S_1$ must address only after $S_2$,we only consider the cases where $S_2$ appears before $S_1$.
By symmetry,exactly half of the total arrangements satisfy this condition.
Therefore,the required number of ways is $\frac{10!}{2}$.

(C) $5$-digit number cannot start with $0$.
Total permutations of $5$ distinct digits taken all at once is $^5P_5 = 5! = 120$.
Numbers starting with $0$ are formed by arranging the remaining $4$ digits in the last $4$ positions,which is $^4P_4 = 4! = 24$.
Therefore,the number of $5$-digit numbers is $120 - 24 = 96$.

(D) The word $DAUGHTER$ contains $8$ distinct letters: $D, A, U, G, H, T, E, R$.
There are $3$ vowels $(A, U, E)$ and $5$ consonants $(D, G, H, T, R)$.
We need to select $2$ vowels out of $3$ and $3$ consonants out of $5$.
The number of ways to select the letters is $^3C_2 \times ^5C_3 = 3 \times 10 = 30$.
Each selection consists of $5$ letters,which can be arranged among themselves in $5!$ ways.
$5! = 120$.
Therefore,the total number of words that can be formed is $30 \times 120 = 3600$.

(C) Each of the $6$ distinct greeting cards can be sent to any of the $4$ people.
Since each card has $4$ choices,the total number of ways to send the cards is $4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^6$.
$4^6 = 4096$.

(B) Given that a lock has $3$ rings and each ring has $8$ digits.
Each ring can be set to any of the $8$ digits independently.
Therefore,the total number of different ways in which the $3$ rings can be rotated is $8 \times 8 \times 8 = 8^3$.
Hence,option $B$ is correct.

(B) We have $m$ rows and $n$ columns,making a total of $mn$ chairs.
We need to seat $m$ students such that no row is vacant. This means each of the $m$ rows must contain exactly one student.
First,we choose one chair out of $n$ available chairs in each of the $m$ rows. Since there are $n$ choices for each of the $m$ rows,the number of ways to select the chairs is $n \times n \times \dots \times n$ ($m$ times) $= n^m$.
Next,the $m$ students can be arranged in these $m$ selected chairs in $m!$ ways.
Therefore,the total number of ways is $n^m \times m!$.

(A) Total number of people = $5 + 4 = 9$.
Total arrangements = $9!$.
To find $\alpha$ (arrangements where one particular man and one particular woman are together),treat them as a single unit.
Now we have $8$ units to arrange,which can be done in $8!$ ways.
Within the unit,the man and woman can be arranged in $2!$ ways.
So,$\alpha = 8! \times 2!$.
To find $\beta$ (arrangements where they are not together),subtract $\alpha$ from the total arrangements:
$\beta = 9! - (8! \times 2!) = 9 \times 8! - 2 \times 8! = 7 \times 8!$.
Now,$\alpha: \beta = (8! \times 2) : (7 \times 8!) = 2 : 7$.

(C) We have $3$ men and $3$ women to be arranged in $6$ seats.
The first and last seats must be occupied by men.
Step $1$: Select $2$ men out of $3$ for the first and last seats and arrange them in $P(3, 2) = 3 \times 2 = 6$ ways.
Step $2$: The remaining $4$ people ($1$ man and $3$ women) can be arranged in the remaining $4$ middle seats in $4! = 24$ ways.
Total number of arrangements $= 6 \times 24 = 144$.

(C) The word $ARRANGEMENT$ has $11$ letters: $A(2), R(2), N(2), E(2), G(1), M(1), T(1)$.
Total arrangements $= \frac{11!}{2! \cdot 2! \cdot 2! \cdot 2!} = \frac{11!}{16}$.
To find the arrangements where two $E$s do not occur together,we subtract the arrangements where they do occur together from the total.
Treating the two $E$s as a single unit $(EE)$,we have $10$ items to arrange: $A(2), R(2), N(2), G(1), M(1), T(1), (EE)(1)$.
Number of arrangements with $E$s together $= \frac{10!}{2! \cdot 2! \cdot 2!} = \frac{10!}{8}$.
Number of arrangements with $E$s not together $= \frac{11!}{16} - \frac{10!}{8} = \frac{11 \cdot 10!}{16} - \frac{2 \cdot 10!}{16} = \frac{9 \cdot 10!}{16} = \frac{9}{16}(10!)$.

(B) The word $VOWEL$ consists of $5$ distinct letters: $V, O, W, E, L$.
There are $2$ vowels: $O$ and $E$.
Since the vowels must always remain together,we treat $(OE)$ as a single unit.
Now,we have $4$ units: $V, W, L, (OE)$.
These $4$ units can be arranged in $4! = 24$ ways.
The $2$ vowels within the unit $(OE)$ can be arranged in $2! = 2$ ways.
Therefore,the total number of words is $24 \times 2 = 48$.

(A) An $8$-digit number is formed by $8$ positions.
For the number to be odd,the last digit (units place) must be one of $\{1, 3, 5, 7, 9\}$,which gives $5$ choices.
The first digit cannot be $0$,so it has $9$ choices $(\{1, 2, 3, 4, 5, 6, 7, 8, 9\})$.
The remaining $6$ positions (from the $2^{nd}$ to the $7^{th}$ digit) can each be filled by any of the $10$ digits $(\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\})$,giving $10^6$ ways.
Therefore,the total number of $8$-digit odd numbers is $9 \times 10^6 \times 5 = 45 \times 10^6$.

(A) four-digit number has four places: $ABCD$.
Given that the first digit $A$ is fixed as $4$,there is only $1$ way to fill this place.
The last digit $D$ can be either $0$ or $5$,so there are $2$ ways to fill this place.
The second digit $B$ can be any digit from $0$ to $9$,so there are $10$ ways.
The third digit $C$ can be any digit from $0$ to $9$,so there are $10$ ways.
Therefore,the total number of such four-digit numbers is $1 \times 10 \times 10 \times 2 = 200$.

(D) The word $KANGAROO$ contains $8$ letters: $K, A, N, G, A, R, O, O$. The letter $A$ appears $2$ times and $O$ appears $2$ times.
Total number of arrangements $= \frac{8!}{2!2!} = \frac{40320}{4} = 10080$.
To find the number of arrangements where $A$'s do not appear together,we subtract the arrangements where $A$'s are together from the total.
Treating the two $A$'s as a single unit $(AA)$,we have $7$ units: $(AA), K, N, G, R, O, O$.
The number of arrangements where $A$'s are together $= \frac{7!}{2!} = \frac{5040}{2} = 2520$.
Therefore,the number of required arrangements $= 10080 - 2520 = 7560$.

(C) To find the number of natural numbers less than $500$ with no repeated digits,we consider $1, 2,$ and $3$-digit numbers separately.
$1$. $1$-digit numbers: The possible numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9$. Total = $9$.
$2$. $2$-digit numbers: The first digit can be any of $9$ digits $(1-9)$ and the second digit can be any of the remaining $9$ digits (including $0$). Total = $9 \times 9 = 81$.
$3$. $3$-digit numbers less than $500$: The hundreds place can be filled by $1, 2, 3,$ or $4$ ($4$ ways). The tens place can be filled by any of the remaining $9$ digits,and the units place by any of the remaining $8$ digits. Total = $4 \times 9 \times 8 = 288$.
Total count = $9 + 81 + 288 = 378$.

(B) The word "$ASSASSINATION$" contains $13$ letters in total: $3$ $A$'s,$4$ $S$'s,$2$ $I$'s,$2$ $N$'s,$1$ $T$,and $1$ $O$.
To keep all $4$ $S$'s together,we treat them as a single block or unit,say $Z$.
Now,we have the letters: $A, A, A, I, I, N, N, T, O, Z$.
This gives us $10$ items to arrange,where $A$ repeats $3$ times,$I$ repeats $2$ times,and $N$ repeats $2$ times.
The number of arrangements is given by $\frac{10!}{3! 2! 2!}$.

(A) Numbers between $10$ and $10000$ can be $2$-digit,$3$-digit,or $4$-digit numbers.
Number of ways to form a $2$-digit number using $5$ distinct digits: $5 \times 4 = 20$.
Number of ways to form a $3$-digit number using $5$ distinct digits: $5 \times 4 \times 3 = 60$.
Number of ways to form a $4$-digit number using $5$ distinct digits: $5 \times 4 \times 3 \times 2 = 120$.
Total numbers formed = $20 + 60 + 120 = 200$.

(B) The total number of ways to arrange $9$ papers is $9!$.
To find the number of ways where the best and worst papers are together,we treat them as a single unit.
This leaves us with $8$ units (the combined pair plus the other $7$ papers),which can be arranged in $8!$ ways.
The best and worst papers can be arranged within their unit in $2!$ ways.
So,the number of ways they are together is $2! \times 8!$.
The number of ways they are never together is the total arrangements minus the arrangements where they are together: $9! - 2! \times 8!$.

(A) The word is $MAXIMA$. Total letters $= 6$.
Vowels are $\{A, I, A\}$ (Total $3$,with $A$ repeating $2$ times).
Consonants are $\{M, X, M\}$ (Total $3$,with $M$ repeating $2$ times).
Number of ways to arrange the vowels together $= \frac{3!}{2!} = 3$.
Number of ways to arrange the consonants together $= \frac{3!}{2!} = 3$.
Since we have two groups (one of vowels and one of consonants),these two groups can be arranged in $2! = 2$ ways.
Total number of arrangements $= 2! \times \left(\frac{3!}{2!}\right) \times \left(\frac{3!}{2!}\right) = 2 \times 3 \times 3 = 18$.

(B) The word '$MOBILE$' has $6$ letters: $M, O, B, I, L, E$.
Consonants are $M, B, L$ ($3$ in total).
Vowels are $O, I, E$ ($3$ in total).
There are $6$ positions in total: $1, 2, 3, 4, 5, 6$.
Odd positions are $1, 3, 5$ ($3$ positions).
We need to place $3$ consonants in $3$ odd positions,which can be done in $3!$ ways.
The remaining $3$ vowels can be placed in the remaining $3$ even positions $(2, 4, 6)$ in $3!$ ways.
Total number of words $= 3! \times 3! = 6 \times 6 = 36$.

(D) The given letters are $A, H, L, R, U$ in alphabetical order.
Total number of letters is $5$.
Words starting with $A$: $4! = 24$ words.
Words starting with $H$: $4! = 24$ words.
Words starting with $L$: $4! = 24$ words.
Now,words starting with $R$:
Words starting with $RA$:
$RAH...$: $RAHLU, RAHUL$ ($2$ words).
So,the rank of $RAHUL$ is $24 + 24 + 24 + 2 = 74$.
Thus,the rank of $RAHUL$ is $74$.

(D) First,arrange the $6$ red balls in a row. The number of ways to arrange $6$ identical red balls is $1$ (since they are identical),but if we treat them as distinct for the sake of permutation logic,it is $6!$. However,usually,balls of the same color are considered identical. Assuming the balls are distinct,the arrangement is $6!$.
There are $7$ possible gaps created by $6$ red balls (including the ends) to place the $6$ black balls: $\_ R \_ R \_ R \_ R \_ R \_ R \_$.
The number of ways to choose $6$ positions out of $7$ is $\binom{7}{6} = 7$.
The number of ways to arrange $6$ black balls in these chosen positions is $6!$.
Thus,the total number of arrangements is $6! \times \binom{7}{6} \times 6! = 6! \times 7 \times 6! = 7 \times 6! \times 6!$.

(C) The word "$ATTAIN$" consists of $6$ letters: $A, T, T, A, I, N$.
To ensure the $T$'s are together,we treat the pair $(TT)$ as a single unit or block.
Now,the letters to be arranged are $(TT), A, A, I, N$.
This gives us $5$ units in total.
Among these $5$ units,the letter $A$ repeats $2$ times.
The number of ways to arrange these $5$ units is given by $\frac{5!}{2!} = \frac{120}{2} = 60$.
Since the two $T$'s are identical,there is only $1$ way to arrange them within their block.
Therefore,the total number of arrangements is $60 \times 1 = 60$.

(C) The word $COMBINATION$ consists of $11$ letters: $C, O, M, B, I, N, A, T, I, O, N$.
The vowels are $O, I, A, I, O$ (total $5$ vowels).
The consonants are $C, M, B, N, N$ (total $6$ consonants).
Since the vowels must come together,we treat the group $(O, I, A, I, O)$ as a single entity.
Now,we have $6$ consonants + $1$ entity = $7$ items to arrange.
The number of arrangements of these $7$ items,where $N$ repeats $2$ times,is $\frac{7!}{2!} = \frac{5040}{2} = 2520$.
Within the vowel group $(O, I, A, I, O)$,there are $5$ letters where $O$ repeats $2$ times and $I$ repeats $2$ times.
The number of arrangements of these vowels is $\frac{5!}{2! \times 2!} = \frac{120}{4} = 30$.
The total number of ways is $2520 \times 30 = 75600$.

(B) First,we arrange the letters of the word $LEADING$ in alphabetical order: $A, D, E, G, I, L, N$.
Words starting with $A$: $6! = 720$.
Words starting with $D$: $6! = 720$.
Words starting with $E$: $6! = 720$.
Total words starting with $A, D, E$ is $720 + 720 + 720 = 2160 > 2017$.
So,the word starts with $E$. Words before $E$ are $720 + 720 = 1440$.
Now,words starting with $EA, ED, EG, EI$: $4 \times 5! = 4 \times 120 = 480$.
Total words up to $EI$: $1440 + 480 = 1920$.
Next,words starting with $ELA, ELD, ELG$: $3 \times 4! = 3 \times 24 = 72$.
Total words up to $ELG$: $1920 + 72 = 1992$.
Next,words starting with $ELI$: $4! = 24$.
Total words up to $ELI$: $1992 + 24 = 2016$.
The $2017^{\text{th}}$ word must be the first word starting with $ELN$.
The remaining letters after $ELN$ are $A, D, G, I$ in alphabetical order.
Thus,the $2017^{\text{th}}$ word is $ELNADGI$.

(B) To form integers greater than $3000$ using digits $\{0, 1, 2, 3, 4, 5\}$ without repetition,we consider numbers with $4, 5,$ and $6$ digits.
$1$. $4$-digit numbers: The first digit must be $3, 4,$ or $5$ ($3$ choices). The remaining $3$ positions can be filled by the remaining $5$ digits in $P(5, 3) = 5 \times 4 \times 3 = 60$ ways. Total = $3 \times 60 = 180$. However,if the first digit is $3$,the number is $3xyz$. If $x=0$,we have $30yz$,which is $> 3000$. If $x > 0$,it is also $> 3000$. So all $180$ are valid.
$2$. $5$-digit numbers: The first digit cannot be $0$ ($5$ choices: $1, 2, 3, 4, 5$). The remaining $4$ positions can be filled by the remaining $5$ digits in $P(5, 4) = 5 \times 4 \times 3 \times 2 = 120$ ways. Total = $5 \times 120 = 600$.
$3$. $6$-digit numbers: The first digit cannot be $0$ ($5$ choices). The remaining $5$ positions can be filled by the remaining $5$ digits in $P(5, 5) = 5! = 120$ ways. Total = $5 \times 120 = 600$.
Summing these: $180 + 600 + 600 = 1380$.

(A) The word is $CAPITAL$. The letters in alphabetical order are $A, A, C, I, L, P, T$.
Total letters = $7$. The letter $A$ repeats $2$ times.
Words starting with $A$: The remaining letters are $A, C, I, L, P, T$ ($6$ letters). Number of arrangements = $\frac{6!}{1!} = 720$.
Words starting with $C$:
Next letter is $A$: Remaining letters are $A, I, L, P, T$ ($5$ letters). Number of arrangements = $5! = 120$.
Next letter is $C$ (not possible).
Next letter is $I$: Remaining letters are $A, A, L, P, T$ ($5$ letters). Number of arrangements = $\frac{5!}{2!} = 60$.
Next letter is $L$: Remaining letters are $A, A, I, P, T$ ($5$ letters). Number of arrangements = $\frac{5!}{2!} = 60$.
Next letter is $P$: Remaining letters are $A, A, I, L, T$ ($5$ letters). Number of arrangements = $\frac{5!}{2!} = 60$.
Next letter is $T$: Remaining letters are $A, A, I, L, P$ ($5$ letters). Number of arrangements = $\frac{5!}{2!} = 60$.
Wait,let us re-evaluate:
Alphabetical order: $A, A, C, I, L, P, T$.
Words starting with $A$: $\frac{6!}{1!} = 720$.
Words starting with $C$:
$CA...$: Remaining letters $A, I, L, P, T$.
$CAA...$: $4! = 24$.
$CAI...$: $4! = 24$.
$CAL...$: $4! = 24$.
$CAP...$:
$CAPA...$: $3! = 6$.
$CAPI...$:
$CAPIA...$: $2! = 2$.
$CAPIL...$:
$CAPILA...$: $1! = 1$.
$CAPITAL$: $1$.
Total rank = $720 + 24 + 24 + 24 + 6 + 2 + 1 + 1 = 802$.

(D) First,we arrange $10$ men in a row. The number of ways to arrange $10$ men is $10!$.
There are $11$ gaps created by $10$ men (including the ends) where $6$ women can be seated so that no two women sit together.
The number of ways to choose $6$ gaps out of $11$ and arrange $6$ women is given by $^{11}P_6$.
Total number of ways $= 10! \times ^{11}P_6 = 10! \times \frac{11!}{(11-6)!} = \frac{10! 11!}{5!}$.

(C) The given digits are $0, 1, 2, 3$. We need to form distinct positive integers using these digits at most once.
Case $I$: $4$-digit integers.
The first place can be filled by any of the $3$ non-zero digits $(1, 2, 3)$. The remaining $3$ places can be filled by the remaining $3$ digits in $3 \times 2 \times 1$ ways.
Number of $4$-digit integers $= 3 \times 3 \times 2 \times 1 = 18$.
Case $II$: $3$-digit integers.
The first place can be filled by $3$ choices $(1, 2, 3)$. The remaining $2$ places can be filled by the remaining $3$ digits in $3 \times 2$ ways.
Number of $3$-digit integers $= 3 \times 3 \times 2 = 18$.
Case $III$: $2$-digit integers.
The first place can be filled by $3$ choices $(1, 2, 3)$. The second place can be filled by the remaining $3$ digits in $3$ ways.
Number of $2$-digit integers $= 3 \times 3 = 9$.
Case $IV$: $1$-digit integers.
The possible integers are $1, 2, 3$.
Number of $1$-digit integers $= 3$.
Total number of distinct positive integers $= 18 + 18 + 9 + 3 = 48$.

(B) Each of the $n$ distinct objects can be placed in either of the $2$ different boxes.
Since each object has $2$ choices,the total number of ways to distribute $n$ distinct objects into $2$ different boxes is $2 \times 2 \times 2 \times \dots \times 2$ ($n$ times).
Therefore,the total number of ways is $2^n$.

(C) Each of the $5$ balls can be placed in any of the $4$ tins independently.
Since each ball has $4$ choices,the total number of ways to place $5$ balls in $4$ tins is $4 \times 4 \times 4 \times 4 \times 4 = 4^5$.

(B) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$. The available digits are ${1, 2, 3, 4, 5, 6, 7}$.
Possible two-digit combinations (tens,units) divisible by $4$ are: $12, 16, 24, 32, 36, 52, 56, 64, 72, 76$.
There are $10$ such pairs.
For each pair,we need to fill the remaining $2$ places (thousands and hundreds) using the remaining $5$ digits.
The number of ways to fill the remaining $2$ places is $P(5, 2) = 5 \times 4 = 20$.
Total numbers $= 10 \times 20 = 200$.