# 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?

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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