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A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if a team has

(i) no girl

(ii) at least 3 girls

(iii) at least one girl and one boy?

1 Answer

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(i) no girl

Total No. No. to be chosen No. of ways to choose
Girls 4 0 4C0
Boys 7 5 7C5

Total number of ways = 4C0 × 7C5

\(=\frac{4!}{0!(4-0)!}\times\frac{7!}{5!(7-5)}\)

\(=\frac{4!}{4!}\times \frac{7!}{5!\space2!}\)

\(=1\times\frac{7\times6}{2}\)

= 21

(ii) at least 3 girls? Since, the team has to consist of at least 3 girls, the team can consist of 

(a) 3 girls and 2 boys 

(b) 4 girls and 1 boy 

(a) 3 girls and 2 boys

Total No. No. to be chosen No. of ways to chosen
Girls 4 3 4C3
Boys 7 2 7C2

Number of ways selecting = 7C× 4C3

=\(\frac{7!}{2!\space5!}\times\frac{4!}{3!\space1!}\)

\(=\frac{7\times6}{2\times1}\times4\)

= 84

(b) 4 girls and 1 boy

Total No. No. to be chosen No. of ways to choose
Girls 4 4 4C4
Boys 7 1 7C1

Number if ways selecting = 7C× 4C4

\(=\frac{7!}{1!\space6!}\times\frac{4!}{4!\space0\!}\)

\(=7\times1\)

= 7

∴ Total number of ways = 84 + 7 = 91 

(iii) at least one girl and one boy? 

A group giving at least one boy and one girl will consist of 

(a) 1 boy and 4 girls 

(b) 2 boys and 3 girls 

(c) 3 boys and 2 girls 

(d) 4 boys and 1 girl

Number of ways of selecting 1 boy and 4 girls = 7C1 × 4C4 = 7 

Number of ways of selecting 2 boys and 3 girls = 7C1 × 4C3 = 84 

Number of ways of selecting 3 boys and 2 girls = 7C2 × 4C2 = 210 

Number of ways of selecting 4 boys and 1 girl = 7C4 × 4C1 = 140 

Hence, total number of ways = 7 + 84 + 210 + 140 = 441 ways.

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