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) ├Ч (No. of ways of choosing 3consonants from 17 consonants)
= (5C2)├Ч (17C3)
By using the formula,
nCr┬а= n!/r!(n тАУ r)!
= 10 ├Ч (17├Ч8├Ч5)
= 10 ├Ч 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 ├Ч 5! = 6800 ├Ч(5├Ч4├Ч3├Ч2├Ч1)
= 6800 ├Ч 120
= 816000
тИ┤┬аThe no. of wordsthat can be formed containing 2 vowels and 3 consonants are 816000.
Question - 2 : - There are 10 persons named P1, P2, P3┬атАж,P10. Out of 10 persons, 5 persons are to be arranged in a line suchthat is each arrangement P1┬аmust occur whereas P4┬аandP5┬аdo 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┬атАжP10)
It is also told that P1┬аshouldbe present and P4┬аand P5┬аshould not be present.
We have to choose 4persons from remaining 7 persons as P1┬аis selected and P4┬аandP5┬аare already removed.
Number of ways =Selecting 4 persons from remaining 7 persons
=┬а7C4
By using the formula,
nCr┬а= n!/r!(n тАУ r)!
7C4┬а= 7! / 4!(7 тАУ 4)!
= 7! / (4! 3!)
= [7├Ч6├Ч5├Ч4!] / (4! 3!)
= [7├Ч6├Ч5] / (3├Ч2├Ч1)
= 7├Ч5
= 35
Now we need to arrangethe chosen 5 people. Since 1 person differs from other.
35 ├Ч 5! = 35 ├Ч(5├Ч4├Ч3├Ч2├Ч1)
= 4200
тИ┤┬аThe 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 тАШMONDAYтАЩ, 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 тАШMONDAYтАЩ
Total letters = 6
(i)┬а4 letters are used ata time
Number of ways = (No.of ways of choosing 4 letters from MONDAY)
= (6C4)
By using the formula,
nCr┬а= n!/r!(n тАУ r)!
6C4┬а= 6! / 4!(6 тАУ 4)!
= 6! / (4! 2!)
= [6├Ч5├Ч4!] / (4! 2!)
= [6├Ч5] / (2├Ч1)
= 3├Ч5
= 15
Now we need to findthe no. of words that can be formed by 4 letters.
15 ├Ч 4! = 15 ├Ч(4├Ч3├Ч2├Ч1)
= 15 ├Ч 24
= 360
тИ┤┬аThe no. of wordsthat can be formed by 4 letters of MONDAY is 360.
(ii)┬аall letters are usedat a time
Total number ofletters in the word тАШMONDAYтАЩ is 6
So, the total no. ofwords that can be formed is 6! = 360
тИ┤┬аThe no. of wordsthat can be formed by 6 letters of MONDAY is 360.
(iii)┬аall letters are usedbut first letter is a vowel ?
In the word тАШMONDAYтАЩthe 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┬а= n!/r!(n тАУ r)!
2C1┬а= 2! / 1!(2 тАУ 1)!
= 2! / (1! 1!)
= (2├Ч1)
= 2
Now we need to findthe no. of words that can be formed by remaining 5 letters.
2 ├Ч 5! = 2 ├Ч(5├Ч4├Ч3├Ч2├Ч1)
= 2 ├Ч 120
= 240
тИ┤┬аThe 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 тАУ 3) things from theremaining (n тАУ 3) things.
Let us find the no. ofways of choosing (r тАУ 3) things.
Number of ways = (No.of ways of choosing (r тАУ 3) things from remaining (n тАУ 3) things)
=┬аn тАУ 3CrтАУ 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 тАУ 2) things which werepresent now. So, the total no. of words that can be formed is (r тАУ 2)!
Total number of wordsformed is:
n тАУ 3Cr тАУ 3┬а├Ч 3! ├Ч (r тАУ 2)!
тИ┤┬аThe no. ofpermutations that can be formed by r things which are chosen from n things inwhich 3 things are always together is┬аn тАУ 3Cr тАУ 3┬а├Ч3! ├Ч (r тАУ 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 тАШINVOLUTEтАЩ
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┬а4C3
And numbre of ways toselect 2 consonants is┬а4C2
Then, number of waysto arrange these 5 letters =┬а4C3┬а├Ч┬а4C2┬а├Ч5!
By using the formula,
nCr┬а= n!/r!(n тАУ r)!
4C3┬а= 4!/3!(4-3)!
= 4!/(3! 1!)
= [4├Ч3!] / 3!
= 4
4C2┬а= 4!/2!(4-2)!
= 4!/(2! 2!)
= [4├Ч3├Ч2!] / (2! 2!)
= [4├Ч3] / (2├Ч1)
= 2 ├Ч 3
= 6
So, by substitutingthe values we get
4C3┬а├Ч┬а4C2┬а├Ч5! = 4 ├Ч 6 ├Ч 5!
= 4 ├Ч 6 ├Ч (5├Ч4├Ч3├Ч2├Ч1)
= 2880
тИ┤┬аThe no. of wordsthat can be formed containing 3 vowels and 2 consonants chosen from тАШINVOLUTEтАЩis 2880.