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

(B) The number of signals that can be generated is the number of arrangements of $5$ flags chosen from $8$ distinct flags.
This is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 8$ and $r = 5$.
Therefore,the number of signals = $^8P_5 = 8 \times 7 \times 6 \times 5 \times 4 = 6720$.

(B) There are $6$ letters and $3$ post boxes.
Each letter can be placed in any of the $3$ post boxes.
For the first letter,there are $3$ choices.
For the second letter,there are $3$ choices.
This continues for all $6$ letters.
Therefore,the total number of ways is $3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$ ways.

(A) The word $MISSISSIPPI$ contains $11$ letters: $M(1), I(4), S(4), P(2)$.
First,we arrange the letters other than $S$,which are $M, I, I, I, I, P, P$. There are $7$ such letters.
The number of ways to arrange these $7$ letters is $\frac{7!}{4! \times 2!} = \frac{7 \times 6 \times 5}{2} = 105$.
Note that $105 = 7 \times 3 \times 5 = 7 \times ^6C_4$.
Now,to ensure no two $S$ are together,we place the $4$ $S$ letters in the gaps created by the $7$ letters. There are $8$ such gaps (including ends).
The number of ways to choose $4$ gaps out of $8$ is $^8C_4$.
Thus,the total number of arrangements is $7 \times ^6C_4 \times ^8C_4$.

(D) The letters of the word $SACHIN$ are $A, C, H, I, N, S$ in alphabetical order.
To find the rank of $SACHIN$,we count the number of words starting with letters before $S$ in the alphabetical list.
The letters before $S$ are $A, C, H, I, N$.
Each of these letters,when placed in the first position,allows the remaining $5$ letters to be arranged in $5!$ ways.
Number of words starting with $A, C, H, I, N = 5 \times 5! = 5 \times 120 = 600$.
The next word in the dictionary order is $SACHIN$ itself.
Therefore,the rank of $SACHIN$ is $600 + 1 = 601$.

(C) The word $EAMCET$ contains $6$ letters: $E, A, M, C, E, T$.
There are $2$ vowels $(E, E)$ and $4$ consonants $(A, M, C, T)$.
Wait,let us re-examine the letters: $E, A, M, C, E, T$.
Vowels: $E, A, E$. Consonants: $M, C, T$.
Total letters = $6$. Vowels = $3$ $(E, A, E)$. Consonants = $3$ $(M, C, T)$.
To ensure no two vowels are together,we arrange the consonants first: $3! = 6$ ways.
This creates $4$ gaps: $\_ C \_ C \_ C \_$.
We need to place $3$ vowels in these $4$ gaps.
The number of ways to choose and arrange the vowels is $^4P_3 = 4 \times 3 \times 2 = 24$.
Since there are two identical vowels $(E, E)$,we divide by $2!$.
Total arrangements = $3! \times \frac{^4P_3}{2!} = 6 \times \frac{24}{2} = 6 \times 12 = 72$.

(C) The word $TRIANGLE$ has $8$ letters: $T, R, I, A, N, G, L, E$.
There are $3$ vowels $(I, A, E)$ and $5$ consonants $(T, R, N, G, L)$.
First,arrange the $5$ consonants in $5!$ ways: $5! = 120$.
These $5$ consonants create $6$ gaps (including ends) where the $3$ vowels can be placed: $\_ C \_ C \_ C \_ C \_ C \_$.
The number of ways to choose and arrange $3$ vowels in $6$ gaps is $^6P_3 = 6 \times 5 \times 4 = 120$.
Total arrangements = $120 \times 120 = 14400$.

(C) The number consists of $2$ pairs of identical digits and $2$ distinct digits.
Step $1$: Select the $2$ digits that will form the pairs from the $4$ available digits: $^4C_2 = 6$ ways.
Step $2$: Select the $2$ remaining distinct digits from the remaining $2$ digits: $^2C_2 = 1$ way.
Step $3$: Arrange these $6$ digits (where $2$ pairs are identical) in a row: $\frac{6!}{2! \times 2!} = \frac{720}{4} = 180$ ways.
Total numbers = $6 \times 1 \times 180 = 1080$.

(A) There are two possible patterns for alternating arrangements:
$1$. $B G B G B G B G B G$
$2$. $G B G B G B G B G B$
In the first pattern,$5$ boys can be arranged in $5!$ ways and $5$ girls can be arranged in $5!$ ways.
Total arrangements for the first pattern $= 5! \times 5!$.
Similarly,for the second pattern,the total arrangements $= 5! \times 5!$.
Total number of arrangements $= 5! \times 5! + 5! \times 5! = 2 \times (5!)^2$.

(A) The word $BANANA$ consists of $6$ letters in total,where $A$ appears $3$ times,$N$ appears $2$ times,and $B$ appears $1$ time.
The number of permutations of $n$ objects where $p$ are of one kind,$q$ are of another kind,and $r$ are of a third kind is given by $\frac{n!}{p!q!r!}$.
Here,$n = 6$,$p = 3$ (for $A$),and $q = 2$ (for $N$).
Total number of permutations $= \frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$.

(D) Given the equation: $4 \times ^{15}P_r = 3 \times ^{16}P_{r-1}$
Using the formula $^{n}P_r = \frac{n!}{(n-r)!}$,we get:
$4 \times \frac{15!}{(15-r)!} = 3 \times \frac{16!}{(16-(r-1))!}$
$4 \times \frac{15!}{(15-r)!} = 3 \times \frac{16 \times 15!}{(17-r)!}$
Divide both sides by $15!$:
$4 \times \frac{1}{(15-r)!} = \frac{48}{(17-r)(16-r)(15-r)!}$
$4 = \frac{48}{(17-r)(16-r)}$
$(17-r)(16-r) = \frac{48}{4} = 12$
$(17-r)(16-r) = 4 \times 3$
Comparing the factors,we have $17-r = 4$ and $16-r = 3$,which gives $r = 13$.

(C) The formula for permutation is $_{n}P_r = \frac{n!}{(n-r)!}$.
Given $_{12}P_r = 1320$.
We know that $_{12}P_r = 12 \times 11 \times 10 \times \dots \times (12-r+1)$.
Let us calculate the product of consecutive integers starting from $12$:
$12 \times 11 = 132$
$12 \times 11 \times 10 = 1320$
Since the product of $3$ terms is $1320$,we have $r = 3$.

(C) The word $EAMCET$ has $6$ letters: $E, A, M, C, E, T$.
There are $2$ vowels $(E, E)$ and $4$ consonants $(A, M, C, T)$. Wait,let us re-examine the letters: $E, A, M, C, E, T$. The vowels are $E, A, E$ ($3$ vowels) and consonants are $M, C, T$ ($3$ consonants).
First,arrange the $3$ consonants $M, C, T$ in $3! = 6$ ways.
These consonants create $4$ gaps (including ends): $\_ M \_ C \_ T \_$.
We need to place the $3$ vowels $(E, A, E)$ in these $4$ gaps such that no two vowels are together.
The number of ways to choose $3$ gaps out of $4$ is $^4C_3 = 4$.
Since there are $2$ identical vowels $(E, E)$,the number of ways to arrange the vowels in the chosen gaps is $\frac{3!}{2!} = 3$.
Total arrangements = $3! \times ^4C_3 \times \frac{3!}{2!} = 6 \times 4 \times 3 = 72$.

(A) The word $COURTESY$ has $8$ distinct letters: $C, O, U, R, T, E, S, Y$.
Since the first letter must be $C$ and the last letter must be $Y$,these two positions are fixed.
We are left with $8 - 2 = 6$ letters to be arranged in the remaining $6$ positions.
The number of ways to arrange $6$ distinct letters is $6!$.

(A) The word $UNIVERSAL$ contains $9$ distinct letters: $U, N, I, V, E, R, S, A, L$.
To form a $3$-letter word,we need to select and arrange $3$ letters out of $9$.
The number of ways to arrange $r$ objects out of $n$ distinct objects is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.
Here,$n = 9$ and $r = 3$.
Number of words $= P(9, 3) = 9 \times 8 \times 7 = 504$.

(C) The letters of the word $RACHIT$ in alphabetical order are: $A, C, H, I, R, T$.
Words starting with $A$: $5! = 120$
Words starting with $C$: $5! = 120$
Words starting with $H$: $5! = 120$
Words starting with $I$: $5! = 120$
After these,the next words start with $R$. The first word starting with $R$ is $RACHIT$.
Therefore,the rank of the word $RACHIT$ is $(4 \times 120) + 1 = 480 + 1 = 481$.

(A) number is divisible by $4$ if the number formed by its last two digits is divisible by $4$.
Using the digits ${1, 2, 3, 4, 5}$,the possible $2$-digit numbers divisible by $4$ are:
$12, 24, 32, 52$.
There are $4$ such pairs.
For each pair,the remaining $3$ positions can be filled by the remaining $3$ digits in $3! = 6$ ways.
Total numbers = $(\text{Number of pairs}) \times (3!) = 4 \times 6 = 24$.

(B) The number of permutations of $n$ distinct objects taken all at a time is given by $n!$.
Here,$n = 3$ (letters $A, B, C$).
Therefore,the number of permutations is $3! = 3 \times 2 \times 1 = 6$.
The possible permutations are: $ABC, ACB, BCA, BAC, CBA, CAB$.

(B) Step $1$: Select $4$ consonants from $6$ consonants: $\binom{6}{4} = \frac{6 \times 5}{2 \times 1} = 15$.
Step $2$: Select $3$ vowels from $5$ vowels: $\binom{5}{3} = \frac{5 \times 4}{2 \times 1} = 10$.
Step $3$: The total number of ways to select the letters is $15 \times 10 = 150$.
Step $4$: These $7$ selected letters can be arranged among themselves in $7!$ ways.
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.
Step $5$: Total number of words = $150 \times 5040 = 756000$.

(A) Numbers between $5000$ and $10000$ are $4$-digit numbers.
The thousands place can be filled by any of the digits from the set $\{5, 6, 7, 8, 9\}$,which gives $5$ choices.
Since no digit is repeated,we have $8$ remaining digits to fill the remaining $3$ places.
The number of ways to fill the remaining $3$ places using $8$ digits is given by $^8P_3$.
Therefore,the total number of such numbers is $5 \times ^8P_3$.

(A) Each of the $5$ prizes can be given to any of the $4$ boys.
Since each prize can be distributed in $4$ ways,the total number of ways to distribute $5$ prizes is $4 \times 4 \times 4 \times 4 \times 4 = 4^5$.
$4^5 = 1024$.

(B) The letters in the word $ARRANGE$ are $A, A, R, R, N, G, E$. Total letters = $7$. The frequency of letters is $A: 2, R: 2, N: 1, G: 1, E: 1$.
To find the rank,we arrange the letters in alphabetical order: $A, A, E, G, N, R, R$.
$1$. Words starting with $A$: Remaining letters are $A, E, G, N, R, R$ ($6$ letters,$R$ repeats twice). Number of words = $\frac{6!}{2!} = 360$.
Wait,the word starts with $A$. Let's fix the first letter as $A$.
Remaining letters: $A, E, G, N, R, R$.
Words starting with $AA$: Remaining $E, G, N, R, R$ ($5$ letters,$R$ repeats twice). Count = $\frac{5!}{2!} = 60$.
Words starting with $AE$: Remaining $A, G, N, R, R$ ($5$ letters,$R$ repeats twice). Count = $\frac{5!}{2!} = 60$.
Words starting with $AG$: Remaining $A, E, N, R, R$ ($5$ letters,$R$ repeats twice). Count = $\frac{5!}{2!} = 60$.
Words starting with $AN$: Remaining $A, E, G, R, R$ ($5$ letters,$R$ repeats twice). Count = $\frac{5!}{2!} = 60$.
Words starting with $AR$: Remaining $A, E, G, N, R$ ($5$ letters). Count = $5! = 120$.
Now,words starting with $ARA$: Remaining $E, G, N, R$ ($4$ letters). Count = $4! = 24$.
Words starting with $ARE$: Remaining $A, G, N, R$ ($4$ letters). Count = $4! = 24$.
Words starting with $ARG$: Remaining $A, E, N, R$ ($4$ letters). Count = $4! = 24$.
Words starting with $ARN$: Remaining $A, E, G, R$ ($4$ letters). Count = $4! = 24$.
Words starting with $ARR$: Remaining $A, A, E, G, N$.
$ARRA...$: $ARRAEGN$ $(1)$,$ARRAG...$ $(2)$,$ARRAN...$ $(3)$.
Summing up the ranks carefully,the rank of $ARRANGE$ is $341$.

(A) The total number of $5$-letter words that can be formed using $10$ distinct letters (with repetition allowed) is $10^5 = 100000$.
The number of $5$-letter words that can be formed using $10$ distinct letters without any repetition is given by the permutation formula $^{10}P_5 = \frac{10!}{(10-5)!} = 10 \times 9 \times 8 \times 7 \times 6 = 30240$.
The number of words with at least one letter repeated is the total number of words minus the number of words with no repeated letters.
Required number of words = $100000 - 30240 = 69760$.

(B) The scholarship for Physics can be awarded to any one of the $3$ applicants,which can be done in $3$ ways.
The scholarship for Mathematics can be awarded to any one of the $5$ applicants,which can be done in $5$ ways.
The scholarship for Chemistry can be awarded to any one of the $4$ applicants,which can be done in $4$ ways.
Since we need to award one scholarship in each subject,we use the fundamental principle of multiplication.
Total number of ways $= 3 \times 5 \times 4 = 60$.

(B) The total number of digits is $5$,where the digit $5$ repeats $2$ times.
The number of permutations is given by the formula:
$N = \frac{n!}{p!} = \frac{5!}{2!} = \frac{120}{2} = 60$.

(C) The word $GARDEN$ has $6$ distinct letters: $G, A, R, D, E, N$.
The vowels in the word are $A$ and $E$.
Total number of arrangements of $6$ letters is $6! = 720$.
In any arrangement,the vowels $A$ and $E$ can appear in $2! = 2$ ways: $(A, E)$ or $(E, A)$.
Since we require the vowels to appear in alphabetical order,we only consider the case where $A$ comes before $E$.
By symmetry,exactly half of the total arrangements will have $A$ before $E$,and the other half will have $E$ before $A$.
Therefore,the number of arrangements where $A$ appears before $E$ is $\frac{6!}{2!} = \frac{720}{2} = 360$.

(A) The special guest must occupy the special chair,which can be done in $1$ way.
Now,there are $8$ remaining chairs and $5$ remaining people to be seated.
The number of ways to arrange $5$ people in $8$ chairs is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.
Here,$n = 8$ and $r = 5$.
Number of ways = $P(8, 5) = \frac{8!}{(8-5)!} = \frac{8!}{3!} = 8 \times 7 \times 6 \times 5 \times 4 = 6720$ ways.

(C) We need to form $4$-digit numbers using the digits ${0, 2, 3, 6, 7, 8}$.
Total permutations of $4$ digits chosen from $6$ is $^6P_4 = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360$.
However,a $4$-digit number cannot have $0$ in the thousands place.
The number of permutations where $0$ is in the thousands place is $^5P_3 = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$.
Therefore,the total number of valid $4$-digit numbers is $360 - 60 = 300$.

(B) Given that $a = {}^{x+2}P_{x+2} = (x+2)!$
$b = {}^{x}P_{11} = \frac{x!}{(x-11)!}$
$c = {}^{x-11}P_{x-11} = (x-11)!$
Given the equation $a = 182bc$:
$(x+2)! = 182 \times \frac{x!}{(x-11)!} \times (x-11)!$
$(x+2)(x+1)x! = 182 \times x!$
$(x+2)(x+1) = 182$
Since $182 = 14 \times 13$,we have $(x+2)(x+1) = 14 \times 13$.
Comparing the factors,$x+2 = 14$ or $x+1 = 13$.
Therefore,$x = 12$.

(B) The total number of books is $n = a + 2 + 3 + 1 = a + 6$.
Since there are $a$ books of type $A$,$2$ identical books of type $B$,$3$ identical books of type $C$,and $1$ book of type $D$,the number of distinct arrangements is given by the formula for permutations of a multiset:
$P = \frac{n!}{n_1! n_2! n_3! n_4!} = \frac{(a + 6)!}{a! 2! 3! 1!}$.
Simplifying this,we get $\frac{(a + 6)!}{a! 2! 3!}$.

(A) The total number of balls is $5 + 4 + 1 = 10$.
Since there are $5$ red balls,$4$ blue balls,and $1$ green ball,the number of distinct arrangements is given by the formula for permutations of a multiset:
$\frac{n!}{n_1! \times n_2! \times n_3!} = \frac{10!}{5! \times 4! \times 1!}$
$= \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times (4 \times 3 \times 2 \times 1) \times 1}$
$= \frac{10 \times 9 \times 8 \times 7 \times 6}{24}$
$= 10 \times 9 \times 7 \times 2 = 1260$.

(A) For a number to be between $3000$ and $4000$,the thousands place must be $3$.
For a number to be divisible by $5$,the units place must be $0$ or $5$. Since $0$ is not in the given set,the units place must be $5$.
Thus,the thousands place is fixed as $3$ and the units place is fixed as $5$.
The remaining two places (hundreds and tens) must be filled using the remaining $4$ digits $(4, 6, 7, 8)$ without repetition.
The number of ways to fill these $2$ places is given by the permutation formula $^4P_2$.
$^4P_2 = \frac{4!}{(4-2)!} = 4 \times 3 = 12$.
Therefore,there are $12$ such numbers.

(D) The word $MATHEMATICS$ contains $11$ letters: $M, M, A, A, T, T, H, E, I, C, S$. There are $3$ pairs of identical letters $(M, A, T)$ and $5$ distinct letters $(H, E, I, C, S)$.
Case $I$: Two letters of one kind and two of another kind.
Number of ways $= ^3C_2 \times \frac{4!}{2!2!} = 3 \times 6 = 18$.
Case $II$: Two letters of one kind and two distinct letters.
Number of ways $= ^3C_1 \times ^7C_2 \times \frac{4!}{2!} = 3 \times 21 \times 12 = 756$.
Case $III$: All four letters are distinct.
Number of ways $= ^8C_4 \times 4! = 70 \times 24 = 1680$.
Total number of arrangements $= 18 + 756 + 1680 = 2454$.

(A) The word $BANANA$ contains $6$ letters: $3$ $A$'s,$2$ $N$'s,and $1$ $B$.
The total number of arrangements is $\frac{6!}{3!2!} = \frac{720}{6 \times 2} = 60$.
To find the arrangements where the two $N$'s appear together,we treat the two $N$'s as a single unit $(NN)$. Now we have $5$ units: $(NN), A, A, A, B$.
The number of such arrangements is $\frac{5!}{3!1!} = \frac{120}{6} = 20$.
The number of arrangements where the two $N$'s do not appear together is the total arrangements minus the arrangements where they are together:
$60 - 20 = 40$.

(B) Let the $5$ animals that cannot enter the $4$ small cages be called 'large animals' and the other $5$ animals be 'small animals'.
There are $6$ large cages and $4$ small cages.
The $5$ large animals must be placed in the $6$ large cages. This can be done in $_6P_5$ ways.
After placing the $5$ large animals,we have $1$ large cage left and $4$ small cages left,totaling $5$ cages for the remaining $5$ small animals.
The $5$ small animals can be placed in these $5$ cages in $5!$ ways.
Therefore,the total number of ways is $_6P_5 \times 5!$.
$_6P_5 = \frac{6!}{1!} = 6 \times 5 \times 4 \times 3 \times 2 = 720$.
$5! = 120$.
Total ways = $720 \times 120 = 86400$.

(A) To form a $4$-digit number using the digits $1, 2, 3, 4, 5, 6, 7$ without repetition,we need to select $4$ digits out of $7$ and arrange them.
The number of ways to select and arrange $4$ digits out of $7$ is given by the permutation formula $^nP_r = \frac{n!}{(n-r)!}$.
Here,$n = 7$ and $r = 4$.
Number of ways = $^7P_4 = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.
Note: The original provided option $480$ is incorrect based on the standard permutation calculation for $4$ distinct digits from $7$.

(C) Given the equation: $\frac{^{56}P_{r+6}}{^{54}P_{r+3}} = 30800$
Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we get:
$\frac{56!}{(56-(r+6))!} \times \frac{(54-(r+3))!}{54!} = 30800$
$\frac{56!}{(50-r)!} \times \frac{(51-r)!}{54!} = 30800$
$\frac{56 \times 55 \times 54!}{(50-r)!} \times \frac{(51-r)(50-r)!}{54!} = 30800$
$56 \times 55 \times (51-r) = 30800$
$3080 \times (51-r) = 30800$
$51-r = \frac{30800}{3080}$
$51-r = 10$
$r = 51 - 10 = 41$

(B) To form an $8$-digit number with distinct digits from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$,the first digit cannot be $0$.
There are $9$ choices for the first digit (any digit from $1$ to $9$).
For the remaining $7$ positions,we need to choose and arrange $7$ digits from the remaining $9$ available digits (including $0$).
The number of ways to fill the remaining $7$ positions is $^9P_7$.
Thus,the total number of $8$-digit numbers is $9 \times ^9P_7$.
Calculating this: $9 \times \frac{9!}{(9-7)!} = 9 \times \frac{9!}{2!} = \frac{9 \times 9!}{2}$.

(D) The total number of $5$-digit numbers that can be formed using the digits $1$ and $2$ is $2^5 = 32$.
The numbers where all digits are the same are $(1, 1, 1, 1, 1)$ and $(2, 2, 2, 2, 2)$.
We are looking for numbers where exactly one digit is different. This means $4$ digits are the same and $1$ digit is different.
Case $1$: Four $1$s and one $2$. The number of arrangements is $\frac{5!}{4!1!} = 5$.
Case $2$: Four $2$s and one $1$. The number of arrangements is $\frac{5!}{4!1!} = 5$.
Total such numbers = $5 + 5 = 10$.

(C) The letters of the word $COCHIN$ are $\{C, C, H, I, N, O\}$.
Total number of permutations = $\frac{6!}{2!} = \frac{720}{2} = 360$.
To find the number of words before $COCHIN$,we arrange the letters in alphabetical order: $C, C, H, I, N, O$.
$1$. Words starting with $C$ (followed by $C$): The remaining letters are $\{H, I, N, O\}$. Number of words = $4! = 24$.
$2$. Words starting with $CH$: The remaining letters are $\{C, I, N, O\}$. Number of words = $4! = 24$.
$3$. Words starting with $CI$: The remaining letters are $\{C, H, N, O\}$. Number of words = $4! = 24$.
$4$. Words starting with $CN$: The remaining letters are $\{C, H, I, O\}$. Number of words = $4! = 24$.
Total words before $COCHIN$ = $24 + 24 + 24 + 24 = 96$.

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

(B) Numbers less than $10000$ can have $1, 2, 3,$ or $4$ digits.
$1$-digit numbers: The digit can be $1, 2, 3, 4, 5, 6, 7$ (since $0$ is not considered a $1$-digit number in this context). Total $= 7$.
$2$-digit numbers: The first place can be filled in $7$ ways $(1-7)$ and the second place in $8$ ways $(0-7)$. Total $= 7 \times 8 = 56$.
$3$-digit numbers: The first place can be filled in $7$ ways,and the second and third places in $8$ ways each. Total $= 7 \times 8 \times 8 = 448$.
$4$-digit numbers: The first place can be filled in $7$ ways,and the remaining three places in $8$ ways each. Total $= 7 \times 8 \times 8 \times 8 = 3584$.
Total numbers $= 7 + 56 + 448 + 3584 = 4095$.

(C) Given the ratio $^nP_4 : ^nP_5 = 1 : 2$.
Using the formula $^nP_r = \frac{n!}{(n-r)!}$,we have:
$\frac{n!}{(n-4)!} \div \frac{n!}{(n-5)!} = \frac{1}{2}$
$\frac{n!}{(n-4)!} \times \frac{(n-5)!}{n!} = \frac{1}{2}$
$\frac{(n-5)!}{(n-4)(n-5)!} = \frac{1}{2}$
$\frac{1}{n-4} = \frac{1}{2}$
$n - 4 = 2$
$n = 6$

(B) number greater than $10$ lakh must have at least $7$ digits.
Since we have exactly $7$ digits $(2, 3, 0, 3, 4, 2, 3)$,any $7$-digit number formed using these will be greater than $10$ lakh,provided the first digit is not $0$.
The total number of permutations of these $7$ digits ($2$ appears twice,$3$ appears thrice,$0$ appears once,$4$ appears once) is given by $\frac{7!}{2! \times 3!} = \frac{5040}{2 \times 6} = 420$.
The number of permutations starting with $0$ (which are $6$-digit numbers) is $\frac{6!}{2! \times 3!} = \frac{720}{2 \times 6} = 60$.
Therefore,the number of $7$-digit numbers greater than $10$ lakh is $420 - 60 = 360$.

(C) We know that the formula for permutation is $^nP_r = \frac{n!}{(n-r)!}$.
Consider the expression $(n - r + 1) \times ^nP_{r - 1}$.
Using the formula,$^nP_{r-1} = \frac{n!}{(n - (r - 1))!} = \frac{n!}{(n - r + 1)!}$.
Substituting this into the expression:
$(n - r + 1) \times \frac{n!}{(n - r + 1)!} = (n - r + 1) \times \frac{n!}{(n - r + 1) \times (n - r)!}$.
Canceling the term $(n - r + 1)$ from the numerator and denominator:
$= \frac{n!}{(n - r)!} = ^nP_r$.
Thus,the correct option is $C$.

(A) The available odd digits are $\{1, 3, 5, 7, 9\}$. There are $5$ such digits.
To form a $6$-digit number using only these $5$ digits such that all digits appear at least once,exactly one digit must be repeated.
Step $1$: Select the digit to be repeated from the $5$ available odd digits in $^5C_1 = 5$ ways.
Step $2$: Arrange these $6$ digits (where one digit is repeated twice) in $\frac{6!}{2!}$ ways.
Total numbers = $5 \times \frac{6!}{2!} = 5 \times \frac{720}{2} = 5 \times 360 = 1800$.

(B) The total number of letters is $6$ $(a, b, c, d, e, f)$. The vowels are $a, e$ ($2$ vowels) and the consonants are $b, c, d, f$ ($4$ consonants).
Total arrangements of $3$ letters from $6$ letters without repetition is $^6P_3 = 6 \times 5 \times 4 = 120$.
Arrangements with no vowels (i.e.,all $3$ letters are consonants) is $^4P_3 = 4 \times 3 \times 2 = 24$.
Therefore,the number of arrangements with at least one vowel is $120 - 24 = 96$.

(B) The word $ARTICLE$ consists of $7$ distinct letters: $A, R, T, I, C, L, E$.
There are $3$ vowels $(A, I, E)$ and $4$ consonants $(R, T, C, L)$.
In a $7$-letter word,the positions are $1, 2, 3, 4, 5, 6, 7$. The even positions are $2, 4, 6$ (total $3$ positions).
The $3$ vowels must occupy these $3$ even positions,which can be done in $^3P_3 = 3! = 6$ ways.
The remaining $4$ consonants must occupy the $4$ odd positions $(1, 3, 5, 7)$,which can be done in $^4P_4 = 4! = 24$ ways.
Total number of words = $3! \times 4! = 6 \times 24 = 144$.