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Q.4.     In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

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A team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ^5C_3 ways.

3 girls can be selected from 4 boys in ^4C_3 ways.

Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected  =^5C_3\times ^4C_3

                                                                                                                                                                         =\frac{5!}{2!3!}\times \frac{4!}{1!3!}

                                                                                                                                                                         =10\times 4=40

Posted by

seema garhwal

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