RD Chapter 17 Combinations Ex 17.3 Solutions
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.