Question

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 three gir

Answer 1

Total Girls =4
Total boys =7
Total members =5
Ways of having no girl
=7c5*4c0=21ways

Ways of atleast one boy and 1 girl
= 7c1*4c4+7c2*4c3+7c3*4c2+7c4*4c1
By solving we get
= 273 ways

Ways of at least 3girls
=7c1*4c4+7c2*4c3
By solving we get
=91 ways
Hope it helps you

Answer 2

Answer:

i) 21

ii) 441

Step-by-step explanation:

i) No girls

      7C5 * 4C0   =   21

ii) 7C1 4C4 + 7C2 4C3 + 7C3 4C2 +7C4 4C1

 = 7!4!/6!4!  +  7!4!/5!2!3!  +  7!4!/4!3!2!2!  +  7!4!/4!3!3!

 =  7 + 84 + 210 + 140

 =  441

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