Q. 194.2( 5 Votes )

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? (ii) at least 3 girls?

Answer :

Given that a group consists of 4 girls and 7 boys.


We need to select a team of 5 members with the following conditions:


i. If a team has no girl


ii. If a team has at least one boy and one girl


iii. If a team has at least 3 girls.


i. Given that we need to select a team of 5 members with no girl present in it out of 4 girls and 7 boys.


Let us assume the no. of ways of selection be N


N = (selecting 5 members out of 7 boys without any girl)


N = 7C5


We know that ,


And also n! = (n)(n – 1)......2.1





N = 21 ways


The no. of ways of selecting 5 members without a girl is 21 ways.


ii. Given that we need to select team of 5 members with at least 1 boy and 1 girl.


Let us assume the no. of ways of selection be N1.


N1 = (Total ways of selecting 5 members out of all 11 members) – (No. of ways of selecting 5 members without any girl)


N1 = (11C5)–(7C5)


We know that ,


And also n! = (n)(n – 1)......2.1





N1 = 462–21


N1 = 441


The no. of ways of selecting 5 members with at least 1 girl and 1 boy is 441.


iii. Given that we need to find the no. of ways to select 5 members with at least 3 girls out of 7 boys and 4 girls.


The following are the possible cases:


i. Selecting 3 girls and 2 boys


ii. Selecting 4 girls and 1 boy


Let us assume the no. of ways of selection be N2.


N2 = (No. of ways of selecting 3 girls and 2 boys out of 7 boys and 4 girls) + (No. of ways of selecting 4 girls and 1 boy out of 7 boys and 4 girls)


N2 = ((4C3) × (7C2)) + ((4C4) × (7C1))


We know that ,


And also n! = (n)(n – 1)......2.1





N2 = (4 × 21) + (1 × 7)


N2 = 84 + 7


N2 = 91


The no. of ways of selecting 5 members with at least 3 girls is 91.


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