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Question: If $(n+1) !=12 \times[(n-1) !]$, find the value of $n .$ Solution: To Find: Value of $n$ Given: $(n+1) !=12 \times[(n-1) !]$ Formula Used: $n !=(n) \times(n-1) \times(n-2) \times(n-3) \ldots \ldots \ldots .3 \times 2 \times 1$ Now, $(n+1) !=12 \times[(n-1) !]$ $\Rightarrow(n+1) \times(n) \times[(n-1) !]=12 \times[(n-1) !]$ $\Rightarrow(n+1) \times(n)=12$ $\Rightarrow n^{2}+n=12$ $\Rightarrow n^{2}+n-12=0$ $\Rightarrow(n-3)(n+4)=0$ $\Rightarrow n=3$ or, $n=-4$ But, n=-4 is not possible ...

In how many different ways can a garland of 16 different flowers be made?

Question: In how many different ways can a garland of 16 different flowers be made? Solution: It is also in the form of a circle, So we need to arrange 16flowers in Circle 16 flowers can be arranged by 15! Now each flower have the same neighbour in the clockwise and anticlockwise arrangement Total number of arrangement are $(15 !) / 2$...

In how many different ways can 20 different pearls be arranged to form a

Question: In how many different ways can 20 different pearls be arranged to form a necklace? Solution: We know that necklace in the form of a circle, So we need to arrange 20 pearls in Circle 20 pearls can be arranged by $19 !$ Now each pearl have the same neighbour in the clockwise and anticlockwise arrangement Total number of arrangement are $(19 !) / 2$...

In how many ways can 8 persons be seated at a round table so that all shall

Question: In how many ways can 8 persons be seated at a round table so that all shall not have the same neighbour in any two arrangement? Solution: By using the formula $(n-1) !$ (Mention in Solution-1) So 8 persons can be arranged by $7 !$ Now each person have the same neighbour in the clockwise and anticlockwise arrangement Total number of arrangement are $(7 !) / 2=2520$...

In how many ways can 11 members of a committee sit at a round table

Question: In how many ways can 11 members of a committee sit at a round table so that the secretary and the joint secretary are always the neighbour of the president? Solution: First assume the president(P), Joint secretary(JS) and secretary(S) to be 1 members(as shown below) So there are 9 members, a number of ways to arrange this 9 people is $8 !$ (The formula used (n-1)!) Now we need to look at the internal arrangement. There are 2 arrangement possible So total number of arrangement are $(8 !...

There are 5 men and 5 ladies to dine at a round table.

Question: There are 5 men and 5 ladies to dine at a round table. In how many ways can they sit so that no ladies are together? Solution: Let first arranged 5 men in the round table by 4! (By using the formula (n-1)! Mention above) Now there are 5 gaps created between 5 men (check the figure) So we arrange 5 ladies in this gap by $5 !$ A total number of ways to arrange 5 men and 5 ladies is $5 ! \times 4 !=2880$...

In how many ways can 6 persons be arranged in

Question: In how many ways can 6 persons be arranged in (i) a line, (ii) a circle? Solution: (i) Let choose 1 person from 6 by ${ }^{6} \mathrm{C}_{1}=6$ and arranged it in line Now choose another person from remaining 5 by ${ }^{5} C_{1}=5$ and arranged it in line Similarly, choose another person from remaining 4 by ${ }^{4} C_{1}=4$ and arranged it in line Similarly, choose another person from remaining 3 by ${ }^{3} C_{1}=3$ and arranged it in line Similarly, choose another person from remain...

There are 4 candidates for the post of a chairman, and one is to be elected

Question: There are 4 candidates for the post of a chairman, and one is to be elected by votes of 5 men. In how many ways can the vote be given? Solution: Let suppose 4 candidates be $C_{1}, C_{2}, C_{3}, C_{4}$ and 5 men be $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}$ Now $M_{1}$ choose any one candidates from four $\left(C_{1}, C_{2}, C_{3}, C_{4}\right)$ and give the vote to him by any 4 ways Similarly, $M_{2}$ choose any one candidates from four $\left(C_{1}, C_{2}, C_{3}, C_{4}\right)$ and give the ...

In how many ways can 4 prizes be given to 3 boys when a boy is eligible

Question: In how many ways can 4 prizes be given to 3 boys when a boy is eligible for all prizes? Solution: Let suppose 4 prizes be $P_{1}, P_{2}, P_{3}, P_{4}$ and 3 boys be $B_{1}, B_{2}, B_{3}$ Now $P_{1}$ can be distributed to 3 boys $\left(B_{1}, B_{2}, B_{3}\right)$ by 3 ways, Similarly, $P_{2}$ can be distributed to 3 boys $\left(B_{1}, B_{2}, B_{3}\right)$ by 3 ways, Similarly, $P_{3}$ can be distributed to 3 boys $\left(B_{1}, B_{2}, B_{3}\right)$ by 3 ways, And $P_{4}$ can be distribut...

How many 4-digit numbers can be formed with the digits

Question: How many 4-digit numbers can be formed with the digits 0, 2, 3, 4, 5 when a digit may be repeated any numbers of time in any arrangement? Solution: Let Suppose 4 digit number as 4 boxes as shown below. First Box is at $1000^{\text {th }}$ place, the Second box is at $100^{\text {th }}$ place, the Third box is at $10^{\text {th }}$ place, and Fourth box is at $1^{\text {st }}$ place. The $1^{\text {st }}$ box can be filled with four numbers $(2,3,4,5)$ if we include 0 in the 1 stbox the...

How many 3-digit numbers are there when a digit may be repeated any

Question: How many 3-digit numbers are there when a digit may be repeated any numbers of time? Solution: Let Suppose 3 digit number as 3 boxes as shown below. First Box is at $100^{\text {th }}$ place, the Second box is at $10^{\text {th }}$ place, and the Third box be at $1^{\text {st }}$ place. To make a 3 digit number $1^{\text {st }}$ box can be filled with nine numbers $(1,2,3,4,5,6,7,8,9)$ if we include 0 in $1^{\text {st }}$ box then it become 2 digit number(i.e 010 is 2 digit number not ...

In how many ways can 3 letters can be posted in 2 letterboxes?

Question: In how many ways can 3 letters can be posted in 2 letterboxes? Solution: Let Suppose Letterbox be $B_{1}, B_{2}$ and letters are $L_{1}, L_{2}, L_{3}$ So $L_{1}$ can be posted in any 2 letterboxes $\left(B_{1}, B_{2}\right)$ by 2 ways Similarly, $L_{2}$ can be posted in any 2 letterbox $\left(B_{1}, B_{2}\right)$ by 2 ways Similarly, $L_{3}$ can be posted in any 2 letterbox $\left(B_{1}, B_{2}\right)$ by 2 ways So total number of ways is $2^{3}=8$...

In how many ways can 5 bananas be distributed among 3 boys,

Question: In how many ways can 5 bananas be distributed among 3 boys, there being no restriction to the number of bananas each boy may get? Solution: As there is 5 banana, So suppose it as $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ And Let the Boy be $\mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}$ So $B_{1}$ can Be distributed to 3 Boys $\left(A_{1}, A_{2}, A_{3}\right)$ by 3 ways, Similarly, $B_{2}, B_{3}, B_{4}, B_{5}$ Can be distributed to 3 Boys by $3^{4}$ So total number of ways is $3^{5}$...

A child has 6 pockets. In how many ways, he can put 5 marbles in his

Question: A child has 6 pockets. In how many ways, he can put 5 marbles in his pocket? Solution: The first marble can be put into the pockets in 6 ways, i.e. Choose 1 Pocket From 6 by ${ }^{6} \mathrm{C}_{1}=6$ Similarly second, third, Fourth, fifth \ Sixth marble. Thus, the number of ways in which the child can put the marbles is $\underline{6^{5}}$...

The letters of the word ‘INDIA’ are arranged as in a dictionary.

Question: The letters of the word INDIA are arranged as in a dictionary. What are the $1^{\text {st }}, 13^{\text {th }}, 49^{\text {th }}$ and $60^{\text {th }}$ words? Solution: Alphabetical arrangement of letters: A,D,I,N $\Rightarrow 1^{\text {st }}$ word: ADIIN To find other words: Case 1: words starting with A Number of words $=\frac{4 !}{2 !}=12$ $\Rightarrow 13^{\text {th }}$ word starts with D and is DAllN Case 2: words starting with D Number of words $=\frac{4 !}{2 !}=12$ Case 3: Words...

How many 6 - digit numbers can be formed by using the digits 4, 5, 0, 3, 4, 5?

Question: How many 6 - digit numbers can be formed by using the digits 4, 5, 0, 3, 4, 5? Solution: To find: number of 6 digit 0 cannot be in the first place because that would make a 5 - digit number Total number of 6 - digit numbers = Total number of number possible - Number of numbers with 0 at the first place Total number of numbers possible $=\frac{6 !}{2 ! 2 !}=180$ Number of numbers with 0 at first place $=\frac{5 !}{2 ! 2 !}=30$ $\Rightarrow$ Number of 6 - digit numbers $=180-30=150$ 150 ...

How many 7 - digit numbers can be formed by using the digits

Question: How many 7 - digit numbers can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? Solution: To find: number of 7 digit 0 cannot be in the first place because that would make a 6 digit number Total number of 7 - digit numbers = Total number of number possible - Number of numbers with 0 at the first place Total number of numbers possible $=\frac{7 !}{3 ! 2 !}=420$ Number of numbers with 0 at first place $=\frac{6 !}{3 ! 2 !}=60$ $\Rightarrow$ Number of 7 - digit numbers $=420-60=360$ 360...

How many numbers can be formed with the digits 2, 3, 4, 5, 4, 3, 2

Question: How many numbers can be formed with the digits 2, 3, 4, 5, 4, 3, 2 so that the odd digits occupy the odd places? Solution: The table shows the places where the odd digits can be placed There are 4 places And 3 odd digits out of which 2 are of the same kind Choose any 3 places out of the four places in ${ }^{4} \mathrm{C}_{3}$ ways $=4$ ways In each way, the 3 digits can be placed in $\frac{3 !}{2 !}$ ways $=3$ ways $\Rightarrow$ Total number of ways in which odd digits occupy odd place...

How many five - digit numbers can be formed with the digits 5, 4, 3, 5, 3?

Question: How many five - digit numbers can be formed with the digits 5, 4, 3, 5, 3? Solution: To find: Number of 5 - digit numbers that can be formed 2 numbers are of 1 kind, and 2 are of another kind Total number of permutations $=\frac{5 !}{2 ! 2 !}=30$ 30 number can be formed...

Find the number of different words by using all the letters of the word,

Question: (i)Find the number of different words by using all the letters of the word, INSTITUTION. In how many of them (ii) are the three T's together (iii) are the first two letters the two $\mathbf{N}^{\prime} \mathbf{s}$ ? Solution: (i) There are 11 letters of which 3 are of 1 kind, 2 are of the 2nd kind ,3 are of the 3rd kind Number of arrangements $=\frac{11 !}{3 ! 2 ! 3 !}=554400$ (ii) Let all the three T's be denoted by a single letter Z New word is INSIUIONZ Number of permutations $=\fra...

In how many ways can the letters of the word ‘INTERMEDIATE’ be arranged

Question: In how many ways can the letters of the word INTERMEDIATE be arranged so that: (i) The vowels always occupy even places? (ii) The relative orders of vowels and consonants do not change? Solution: (i) There are 6 even places and 6 vowels out of which 2 are of 1 kind, 3 are of the $2^{\text {nd }}$ kind The vowels can be arranged in $\frac{6 !}{2 ! 3 !}=60$ There are 6 consonants out of which 2 is of one kind Number of permutations $=\frac{6 !}{2 !}=360$ $\Rightarrow$ Total number of wor...

How many arrangements can be made by using all the letters of the word

Question: (i)How many arrangements can be made by using all the letters of the word MATHEMATICS? (ii) How many of them begin with C? (iii) How many of them begin with T? Solution: (i) There are 11 letters of which 2 are of 1 kind, 2 are of another kind, 2 are of the 3rd kind Total number of arrangements $=\frac{11 !}{2 ! 2 ! 2 !}=4989600$ (ii) There are 10 spaces to be filled by 10 letters of which 2 are of 3 different kinds Number of arrangements $=\frac{10 !}{2 ! 2 ! 2 !}=453600$ (iii) There a...

In how many ways can the letters of the word ‘ASSASSINATION’

Question: In how many ways can the letters of the word ASSASSINATION be arranged so that all Ss are together? Solution: To find: number of ways letters can be arranged such that all Ss are together Let all Ss be represented by a single letter Z New word is AAINATIONZ Number of arrangements $=\frac{10 !}{3 ! 2 ! 2 !}=151200$ Letters can be arranged in 151200 ways...

How many different words can be formed with the letters of the word

Question: How many different words can be formed with the letters of the word CAPTAIN? In how many of these C and T are never together? Solution: To find: number of words such that C and T are never together Number of words where C and T are never the together = Total numbers of words - Number of words where C and T are together Total number of words $=\frac{7 !}{2 !}=2520$ Let C and T be denoted by a single letter Z $\Rightarrow$ New word is APAINZ This can be permuted in $\frac{6 !}{2 !}=360$ ...

In how many ways can the letters of the word ‘PARALLEL’

Question: In how many ways can the letters of the word PARALLEL be arranged so that all Ls do not come together? Solution: To find: number of words where L do not come together Let the three Ls be treated as a single letter say Z Number of words with L not the together = Total number of words - Words with Ls together The new word is PARAEZ Total number of words $=\frac{8 !}{2 ! 3 !}=3360$ Words with $L$ together $=6 !=720$ ⇒ Words with L, not together = 3360 720 = 2640 There are 2640 words where...

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