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

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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} ?

Answer:

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

Answer:

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} .

Answer:

(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} ?

Answer:

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?

Answer:

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} .

Answer:

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?

Answer:

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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