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RD Chapter 17- Combinations Ex-17.3 Interview Questions Answers

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

Answer 1 :

Given:

Total number of vowels= 5

Total number ofconsonants = 17

Number of ways = (No.of ways of choosing 2 vowels from 5 vowels) × (No. of ways of choosing 3consonants from 17 consonants)

= (5C2)× (17C3)

By using the formula,

nCr = n!/r!(n – r)!

= 10 × (17×8×5)

= 10 × 680

= 6800

Now we need to findthe no. of words that can be formed by 2 vowels and 3 consonants.

The arrangement issimilar to that of arranging n people in n places which are n! Ways to arrange.So, the total no. of words that can be formed is 5!

So, 6800 × 5! = 6800 ×(5×4×3×2×1)

= 6800 × 120

= 816000

 The no. of wordsthat can be formed containing 2 vowels and 3 consonants are 816000.

Question 2 :

There are 10 persons named P1, P2, P3 …,P10. Out of 10 persons, 5 persons are to be arranged in a line suchthat is each arrangement P1 must occur whereas P4 andP5 do not occur. Find the number of such possible arrangements.

Answer 2 :

Given:

Total persons = 10

Number of persons tobe selected = 5 from 10 persons (P1, P2, P3 …P10)

It is also told that P1 shouldbe present and P4 and P5 should not be present.

We have to choose 4persons from remaining 7 persons as P1 is selected and P4 andP5 are already removed.

Number of ways =Selecting 4 persons from remaining 7 persons

7C4

By using the formula,

nCr = n!/r!(n – r)!

7C4 = 7! / 4!(7 – 4)!

= 7! / (4! 3!)

= [7×6×5×4!] / (4! 3!)

= [7×6×5] / (3×2×1)

= 7×5

= 35

Now we need to arrangethe chosen 5 people. Since 1 person differs from other.

35 × 5! = 35 ×(5×4×3×2×1)

= 4200

 The total no. ofpossible arrangement can be done is 4200.

Question 3 :
How many words, with or without meaning can be formed from the letters of the word ‘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 3 :

Given:

The word ‘MONDAY’

Total letters = 6

(i) 4 letters are used ata time

Number of ways = (No.of ways of choosing 4 letters from MONDAY)

= (6C4)

By using the formula,

nCr = n!/r!(n – r)!

6C4 = 6! / 4!(6 – 4)!

= 6! / (4! 2!)

= [6×5×4!] / (4! 2!)

= [6×5] / (2×1)

= 3×5

= 15

Now we need to findthe no. of words that can be formed by 4 letters.

15 × 4! = 15 ×(4×3×2×1)

= 15 × 24

= 360

 The no. of wordsthat can be formed by 4 letters of MONDAY is 360.

(ii) all letters are usedat a time

Total number ofletters in the word ‘MONDAY’ is 6

So, the total no. ofwords that can be formed is 6! = 360

 The no. of wordsthat can be formed by 6 letters of MONDAY is 360.

(iii) all letters are usedbut first letter is a vowel ?

In the word ‘MONDAY’the vowels are O and A. We need to choose one vowel from these 2 vowels for thefirst place of the word.

So,

Number of ways = (No.of ways of choosing a vowel from 2 vowels)

= (2C1)

By using the formula,

nCr = n!/r!(n – r)!

2C1 = 2! / 1!(2 – 1)!

= 2! / (1! 1!)

= (2×1)

= 2

Now we need to findthe no. of words that can be formed by remaining 5 letters.

2 × 5! = 2 ×(5×4×3×2×1)

= 2 × 120

= 240

 The no. of wordsthat can be formed by all letters of MONDAY in which the first letter is avowel is 240.

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

Answer 4 :

Here, it is clear that3 things are already selected and we need to choose (r – 3) things from theremaining (n – 3) things.

Let us find the no. ofways of choosing (r – 3) things.

Number of ways = (No.of ways of choosing (r – 3) things from remaining (n – 3) things)

n – 3Cr– 3

Now we need to findthe no. of permutations than can be formed using 3 things which are together.So, the total no. of words that can be formed is 3!

Now let us assume thetogether things as a single thing this gives us total (r – 2) things which werepresent now. So, the total no. of words that can be formed is (r – 2)!

Total number of wordsformed is:

n – 3Cr – 3 × 3! × (r – 2)!

 The no. ofpermutations that can be formed by r things which are chosen from n things inwhich 3 things are always together is n – 3Cr – 3 ×3! × (r – 2)!

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

Answer 5 :

Given:

The word ‘INVOLUTE’

Total number ofletters = 8

Total vowels are = I,O, U, E

Total consonants = N,V, L, T

So number of ways toselect 3 vowels is 4C3

And numbre of ways toselect 2 consonants is 4C2

Then, number of waysto arrange these 5 letters = 4C3 × 4C2 ×5!

By using the formula,

nCr = n!/r!(n – r)!

4C3 = 4!/3!(4-3)!

= 4!/(3! 1!)

= [4×3!] / 3!

= 4

4C2 = 4!/2!(4-2)!

= 4!/(2! 2!)

= [4×3×2!] / (2! 2!)

= [4×3] / (2×1)

= 2 × 3

= 6

So, by substitutingthe values we get

4C3 × 4C2 ×5! = 4 × 6 × 5!

= 4 × 6 × (5×4×3×2×1)

= 2880

 The no. of wordsthat can be formed containing 3 vowels and 2 consonants chosen from ‘INVOLUTE’is 2880.


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RD Chapter 17- Combinations Ex-17.3 Contributors

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