Question

A committee of 5 students is to be chosen among 6 boys and 5 girls find the number of ways in which the committee should be selected provided that at it should have at least one boy and one girl​

Answer 1

The number of ways in which the committee should be selected provided that it should have at least one boy and one girl is 455.

Step-by-step explanation:

Total no. of students required to be chosen = 5 students

No. of boys = 6  

No. of girls = 5

We have to choose a committee which should consists of at least one boy and one girl, so,

No. of ways 1 boy & 4 girls can be selected = ⁶C₁ * ⁵C₄

No. of ways 2 boys & 3 girls can be selected = ⁶C₂ * ⁵C₃

No. of ways 3 boys & 2 girls can be selected = ⁶C₃ * ⁵C₂

No. of ways 4 boys & 1 girl can be selected = ⁶C₄ * ⁵C₁

Thus,

The total no. of ways in which the committee should be selected, provided that it should have at least one boy and one girl is,

= [⁶C₁ * ⁵C₄] + [⁶C₂ * ⁵C₃] + [⁶C₃ * ⁵C₂ ] + [⁶C₄ * ⁵C₁]

= [$\frac{6!}{5!*1!} * \frac{5!}{1!*4!}$] + [$\frac{6!}{4!*2!} *\frac{5!}{2!*3!}$]+ [$\frac{6!}{3!*3!} * \frac{5!}{3!*2!}$] + [$\frac{6!}{2!*4!} *\frac{5!}{4!*1!}$]

= [6*5] + [15*10] + [20*10] + [15*5]

= 30 + 150 + 200 + 75

= 455

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Answer 2

Answer:

455

Step-by-step explanation:

Required number of ways

= Total possible ways – (ways of selection of girls + ways of selection of boys)only

= ⁶⁺⁵C₅ – (⁵C₅ + ⁶C₅)

= ¹¹C₅ – (5!/5! + 6!/5!)

= 462 – (1 + 6)

= 455

Alternative Solution:

Required number of ways

= C(6, 1) C(5, 4) + C(6, 2) C(5, 3) + C(6, 3) C(5, 2) + C(6, 4) C(5, 1)

= 30 + 150 + 200 + 75

= 455