In the previous section, we completed a discussion on combinations. We saw some solved examples also. In this section we will see some miscellaneous examples.
Solved example 7.21
How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?
Solution:
1. The word INVOLUTE has,
♦ 4 vowels: I, O, U, E
♦ 4 consonants: N, V, L, T
2. Each word must have 3 vowels and 2 consonants. That means, each word will have 5 letters. So let there be 5 boxes.
♦ The first 3 boxes can be considered as one unit.
♦ The remaining two boxes can be considered as another unit.
• This is shown in the fig.7.10 below:
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| Fig.7.10 |
• There are 4 vowels. So this unit can be filled in in 4C3 ways.
4. The remaining unit is for consonants.
• There are 4 consonants. So this unit can be filled in in 4C2 ways.
5. So by applying the multiplication principle, the two units can be filled in:
4C3 × 4C2 ways
6. Consider the following two arrangements:
♦ IOUNV
♦ IUONV
• These are two different permutations of the letters I, U, O, N, V.
♦ But same combination of those letters.
♦ Different combinations will be counted only once
♦ So in this problem, we want no. of permutations, not no. of combinations
7. So we must modify the result obtained in (5). It can be done in 2 steps:
(i) In each of the 4C3 × 4C2 combinations, the five letters can be arranged in 5! ways.
(ii) So the number of permutations = 4C3 × 4C2 × 5! = 2880 ways
Solved example 7.22
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ?
(iii) at least 3 girls ?
Solution:
◼ This is not a permutation problem. It is a combination problem. Each arrangement should contain the specified numbers. The order is not important.
Part (i):
1. Let there be 5 boxes.
All those 5 boxes should be filled with B. None of those boxes must contain G
2. That means, we must use only the seven boys.
5 boys can be selected from 7 boys in 7C5 = 21 ways.
Part (ii):
1. At least one boy and one girl can be selected in the following ways:
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl
2. Consider the arrangement in (a)
♦ 1 boy can be selected from 7 boys in 7C1 ways.
♦ 4 girls can be selected from 4 girls in 4C4 ways.
• So 1 boy and 4 girls can be selected in 7C1 × 4C4 = 7 ways.
3. Consider the arrangement in (b)
♦ 2 boys can be selected from 7 boys in 7C2 ways.
♦ 3 girls can be selected from 4 girls in 4C3 ways.
• So 2 boys and 3 girls can be selected in 7C2 × 4C3 = 84 ways.
4. Consider the arrangement in (c)
♦ 3 boys can be selected from 7 boys in 7C3 ways.
♦ 2 girls can be selected from 4 girls in 4C2 ways.
• So 3 boys and 2 girls can be selected in 7C3 × 4C2 = 210 ways.
5. Consider the arrangement in (d)
♦ 4 boys can be selected from 7 boys in 7C4 ways.
♦ 1 girl can be selected from 4 girls in 4C1 ways.
• So 4 boys and 1 girl can be selected in 7C4 × 4C1 = 140 ways.
6. So total number of ways for selecting at least one boy and one girl is:
(7 + 84 + 210 +140) = 441 ways.
Part (iii):
1. At least 3 girls can be selected in the following ways:
(a) 3 girls and 2 boys
(b) 4 girls and 1 boy
(more than 4 girls is not possible because, the maximum number of girls in the given group is 4)
2. Consider the arrangement in (a)
♦ 3 girls can be selected from 4 girls in 4C3 ways.
♦ 2 boys can be selected from 7 boys in 7C2 ways.
• So 3 girls and 2 boys can be selected in 4C3 × 7C2 = 84 ways.
3. Consider the arrangement in (b)
♦ 4 girls can be selected from 4 girls in 4C4 ways.
♦ 1 boy can be selected from 7 boys in 7C1 ways.
• So 4 girls and 1 boy can be selected in 4C4 × 7C1 = 7 ways.
4. So total number of ways for selecting at least three girls is:
(84 + 7) = 91 ways.
Solved example 7.23
Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the 50th word?
Solution:
◼ This is a permutation problem. The order is important. Each order will give a different word.
Part (i):
• There are 5 letters in the word AGAIN. The letter A appears twice.
• So the number of words = $\frac{5!}{2!}$ = 60
Part (ii):
1. Let us find the number of words beginning with A:
(i) Let there be 5 boxes.
(ii) The first box is fixed with A
(iii) The remaining 4 boxes can be filled in 4! ways.
(iv) So the number of words beginning with A = (1 × 4!) = 24
2. In the word AGAIN, the letter coming after A in the alphabetical order is G.
• Let us find the number of words beginning with G:
(i) Let there be 5 boxes.
(ii) The first box is fixed with G
(iii) The remaining 4 boxes can be filled in $\frac{4!}{2!}$ = 12 ways.
(iv) So the number of words beginning with G = (1 × 4!) = 12
3. In the word AGAIN, the letter coming after G in the alphabetical order is I.
• Let us find the number of words beginning with I:
(i) Let there be 5 boxes.
(ii) The first box is fixed with I
(iii) The remaining 4 boxes can be filled in $\frac{4!}{2!}$ = 12 ways.
(iv) So the number of words beginning with I = (1 × 4!) = 12
4. From the above steps, we get:
Number of words beginning with A, G and I = (24+12+12) =48
5. In the word AGAIN, the letter coming after I in the alphabetical order is N.
• So the 49th word will begin with N.
6. The letters after N in the alphabetical order are: A, A, G, I
• So the 49th word is NAAGI
7. The next alphabetical order is: N, A, A, I, G
• So the 50th word is NAAIG
Solved example 7.24
How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?
Solution:
◼ This is a permutation problem. The order is important. Each order will give a different number.
1. There are seven digits in the given number 1000000.
• So we must use all the given seven digits to form numbers. Then only we will get numbers greater than 1000000
2. Among the given digits, ‘2’ is present three times. Also ‘4’ is present two times.
• So the number of seven digit numbers = $\frac{7!}{(3!)(2!)}$ = 420
3. Out of these numbers, some will be beginning with ‘0’
• Number of numbers beginning with ‘0’ is: $\frac{6!}{(3!)(2!)}$ = 60
4. So the actual number of numbers greater than 1000000 is:
(420 - 60) = 360
Solved example 7.25
In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?
Solution:
◼ This is a permutation problem. The order is important. Each order will give a different arrangement.
1. Consider the following arrangement:
XGXGXGXGXGX
2. Boys can be seated at any of the 'X' marks.
• There are six 'X' marks. So there are six seats available for boys.
• Three boys can sit in six seats in 6P3 = 120 ways.
3. Five girls can be seated in 5! = 120 ways.
4. So the required arrangement can be achieved in (120 × 120) = 14400 ways
The link below gives some more solved examples:
Miscellaneous Exercise on chapter 7
We have completed a discussion on permutations and combinations. In the next chapter, we will see Binomial theorem.
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