Question 1: How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?

$2$ vowels and $3$ consonants can be formed with $5$ vowels and $17$ consonants can be chosen in $^{5} \rm C_{2} \times ^{17} \rm C_{3}$

The five letters can be arranged in $5!$ ways.

Hence the total number of words that can be formed

$= (^{5} \rm C_{2} \times ^{17} \rm C_{3}) \times 5! =$ $\frac{5!}{2!3!}$ $\times$ $\frac{17!}{3! 14!}$ $\times 5!= 6800 \times 120 = 81600$

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Question 2: There are 10 persons named $P_1, P_2, P_3, \ldots P_{10}$ Out of 10 persons, 5 persons are to be arranged in a line such that is each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.

We have to arrange $5$ persons out of $10$ persons in such a way that $P_1$ is always there and $P_4$ and $P_5$ are not selected.

So basically we have to select $4$ persons out of $7$ persons. Hence the number of combinations $= ^{7} \rm C_{4} =$ $\frac{7!}{4! 3!}$ $=$ $\frac{7 \times 6 \times 5}{3 \times 2 \times 1}$ $= 35$

Once we have the persons selected, they can be arranged in $5!$ ways.

Therefore the number of ways $5$ persons can be arranged in a line such that is each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur out of given $10$ persons $= 35 \times 5! = 4200$

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Question 3: How many words, with or without meaning can be formed from the letters of the word $\text{MONDAY}$, assuming that no letter is repeated, if (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel?

i) Number of 4 lettered words that can be formed from the letters of the word $\text{MONDAY}$, assuming that no letter is repeated and if 4 letters are used at a time $= ^{6} \rm C_{4} \times 4! = 360$

ii) Number of 4 lettered words that can be formed from the letters of the word $\text{MONDAY}$, assuming that no letter is repeated and all letters are used at a time $= 6! = 720$

iii) The first place can be filled in 2 ways ( either $\text{A or O}$ ). Therefore one vowel can be chosen from 2 vowels in $^{2} \rm C_{1} = 2$ ways.

Remaining 5 letters can be chosen in $^{5} \rm C_{5} = 1$ way

Therefore the number of words that can be formed $= 2 \times 1 \times 5! = 240$

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Question 4: Find the number of permutations of $n$ distinct things taken $r$ together, in which 3 particular things must occur together.

The number of combination of $n$ distinct things taken r together $= ^{n} \rm C_{r}$

If three things come together $= ^{n-3} \rm C_{r-3}$

Number of arrangements of three things $=3!$

Number of arrangements of $(r-3+1)$ objects $= (r-2)!$

So, total possible ways $= ^{n-3} \rm C_{r-3} \times (r-2)! 3!$

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Question 5: How many words each of 3 vowels and 2 consonants can be formed from the letters of the word $\text{INVOLUTE}$?

Given word $\text{INVOLUTE}$

Number of letters $= 8$

Vowels $= I, O, U, E$

Consonants $= N, V, L, T$

Number of ways to select $3$ vowels $= ^{4} \rm C_{3}$

Number of ways to select $2$ consonants $= ^{4} \rm C_{2}$

Therefore the number of ways to arrange these $5$ letters $= ^{4} \rm C_{3} \times ^{4} \rm C_{2} \times 5! = 4 \times ( 2 \times 3) \times 120 = 2880$

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Question 6: Find the number of permutations of $n$ different things taken $r$ at a time such that $2$ specified things occur together

The number of combination of $n$ distinct things taken $r$ together $= ^{n} \rm C_{r}$

If three things come together $= ^{n-2} \rm C_{r-2}$

Number of arrangements of two things $=2!$

Number of arrangements of $(r-2+1)$ objects $= (r-1)!$

So, total possible ways $= ^{n-2} \rm C_{r-2} \times (r-1)! 2!$

$=$ $\frac{(n-2)!}{(r-2)! [(n-2)-(r-2)]!}$ $(r-1)(r-2)! \cdot 2$

$=$ $\frac{(n-2)!}{[(n-2)-(r-2)]!}$ $(r-1) \cdot 2$

$= 2 \cdot (r-1) ^{n-2} \rm P_{r-2}$

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Question 7: Find the number of ways in which : (a) a selection (b) an arrangement, of four letters can be made from the letters of the word $\text{PROPORTION}$.

(a) Given word: $\text{PROPORTION}$.

Total number of letters $= 10$

Number of $\text{P} =2$  Number of $\text{R} =2$  Number of $\text{O} =3$

Number of $\text{T} =1$  Number of $\text{I} =1$  Number of $\text{N} =1$

The four letter words may consist of

i) 3 alike letters and $1$ distinct letter

Number of ways to select these letters $= 1 \times ^{5} \rm C_{1} = 5$

ii) $2$ alike letters of one kind and $2$ alike letters of second kind

There are three pairs of letters where there are more than one letters. We need to select any $2$ of the $3$ letters.

Number of ways to select these letters $= ^{3} \rm C_{2} = 3$

iii) $2$ alike letters and $2$ distinct letters

Number of ways to select these letters $= ^{3} \rm C_{1} \times ^{5} \rm C_{2} = 3 \times 10 = 30$

iv) all different letters

Number of ways to select these letters $= ^{6} \rm C_{4} = 15$

Therefore the number of ways in which : (a) a selection (b) an arrangement, of four letters can be made from the letters of the word $\text{PROPORTION} = 5+ 3+ 30 +15 = 53$

(b) 4 letter word may consist of:

i) 3 alike letters and $1$ distinct letter

Number of arrangements with 3 alike and one distinct $= 1 \times ^{5} \rm C_{1} \times$ $\frac{4!}{3! 1!}$ $= 20$

ii) $2$ alike letters of one kind and $2$ alike letters of second kind

Number of arrangements with $2$ alike letters of one kind and $2$ alike letters of second kind $= ^{3} \rm C_{2} \times$ $\frac{4!}{2! 2!}$ $= 3 \times 6 = 18$

iii) $2$ alike letters and $2$ distinct letters

Number of arrangements with two letter are same and 2 are of different kind $= ^{3} \rm C_{1} \times ^{5} \rm C_{2} \times$ $\frac{4!}{2! 1! 1!}$ $= 3 \times 10 \times 12 = 360$

iv) all different letters

Number of arrangements for 4 distinct letters $= ^{6} \rm C_{4} \times 4! = 15 \times 24 = 360$

Therefore the total number of arrangements possible $= 20 + 18 + 360 + 360 = 758$

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Question 8: How many words can be formed by taking 4 letters at a time from the letters of the word $\text{MORADABAD}$?

Given word: $\text{MORADABAD}$.

Total number of letters $= 10$

Number of $\text{M} =1$  Number of $\text{O} =1$  Number of $\text{R} =1$

Number of $\text{A} =3$  Number of $\text{D} =2$  Number of $\text{B} =1$

i) 3 alike letters and $1$ distinct letter

Number of arrangements with 3 alike and one distinct $= 1 \times ^{5} \rm C_{1} = 5 \times$ $\frac{4!}{3! 1!}$ $= 20$

ii) $2$ alike letters of one kind and $2$ alike letters of second kind

Number of arrangements with $2$ alike letters of one kind and $2$ alike letters of second kind $= ^{2} \rm C_{2} \times ^{5} \rm C_{2} \times$ $\frac{4!}{2! 2!}$ $= 6$

iii) $2$ alike letters and $2$ distinct letters

Number of arrangements with two letter are same and 2 are of different kind $= ^{2} \rm C_{1} \times ^{5} \rm C_{2} \times$ $\frac{4!}{2! }$ $= 2 \times 10 \times 12 = 240$

iv) all different letters

Number of arrangements for 4 distinct letters $= ^{6} \rm C_{4} \times 4! = 15 \times 24 = 360$

Therefore the total number of arrangements possible $= 20 + 5 + 240 + 360 = 626$

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Question 9: A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?

15 people can be accommodated on the table in $= ^{21} \rm C_{15} \times (15-1)! = ^{21} \rm C_{15} \times 14!$

The remaining 6 people can be arranged in $= ^{6} \rm C_{6} \times (6-1)! = 5!$

Hence the total number of ways people can be arranged $= ^{21} \rm C_{15} \times 14! \times 5!$

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Question 10: Find the number of combinations and permutations of 4 letters taken from the word $\text{EXAMINATION}$.

Given word: $\text{EXAMINATION}$.

Total number of letters $= 11$

Number of $\text{A} =2$  Number of $\text{I} =2$  Number of $\text{N} =2$

All other letters are not repeated.

i) $2$ alike letters of one kind and $2$ alike letters of second kind

Number of arrangements with $2$ alike letters of one kind and $2$ alike letters of second kind $= ^{3} \rm C_{2} \times$ $\frac{4!}{2! 2!}$ $= 18$

ii) $2$ alike letters and $2$ distinct letters

Number of arrangements with two letter are same and 2 are of different kind $= ^{3} \rm C_{1} \times ^{7} \rm C_{2} \times$ $\frac{4!}{2! }$ $= 756$

iii) all different letters

Number of arrangements for 4 distinct letters $= ^{8} \rm C_{4} \times 4! = 8 \times 7 \times 6 \times 5 = 1680$

Therefore the total number of arrangements possible $= 18+756+1680 = 2454$

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Question 11: A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on the other side. In how many ways can they be seated?

We have to arranged for $16$ persons along two sides of a long table with $8$ chairs on each side.

$4$ persons wish to sit on one particular side $A$ and $2$ on the other side $B$.

Out of the $10$ people left, $4$ people can be seated on side $A$ in $^{10} \rm C_{4}$ ways.

And from the remaining $6$ people can be selected for side $B$ in $^{6} \rm C_{6}$ ways.

Hence the number of selections $= ^{10} \rm C_{4} \times ^{6} \rm C_{6}$

Now $8$ people on each side can be arranged in $8!$ways.

Therefore the total number of ways in which the people can be seated  $= ^{10} \rm C_{4} \times ^{6} \rm C_{6} \times 8! \times 8! = ^{10} \rm C_{4} \times (8!)^2$

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