Question

There are 6 boys and 6 girls. The number of ways to choose at least 2 persons so that the combination includes at least a boy and a girl is
$A.{\text{ }}{{\text{2}}^{12}} - {2^7} - 35 \\\ B.{\text{ 36}} \\\ {\text{C}}{\text{. }}{{\text{2}}^{12}} - 13 \\\ D.{\text{ }}{{\text{2}}^{12}} - {2^7} + 1 \\\$

Answer 1

Split this question into two parts one is to find the number of ways to choose at least 2 persons other is to find combination of including at least one boy and one girl add them. So this problem is based on the permutations and combinations. We will use the concept of both to get the answer.

Complete step-by-step answer:
As we know the number of ways in which $r$items can be selected from $n$items is given by $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Number of ways a child can be selected = 2
Total number of ways of selecting all 12 children =${2^{12}}$
No. of ways of selecting at most 1 child ${ = ^{12}}{C_0}{ + ^{12}}{C_1}$
[using compliment]
No. of ways of selecting at least 2 children = (Total number of ways of selecting 12 children) - (no. of way of selecting at most 1 child)
$= {2^{12}}{ - ^{12}}{C_0}{ - ^{12}}{C_1}$
No. of ways of not selecting at least 1 girl and at least 1 boy $= {(^6}{C_2}{ + ^6}{C_3}{ + ^6}{C_4}{ + ^6}{C_5}{ + ^6}{C_6}) + {(^6}{C_2}{ + ^6}{C_3}{ + ^6}{C_4}{ + ^6}{C_5}{ + ^6}{C_6}) \\\ = 2 \times {(^6}{C_2}{ + ^6}{C_3}{ + ^6}{C_4}{ + ^6}{C_5}{ + ^6}{C_6}) \\\ = 2 \times (15 + 20 + 15 + 6 + 1) \\\ = 2(57) \\\ {\text{rewrite 57 as ((2}} \times {\text{2}} \times {\text{2}} \times {\text{2}} \times {\text{2}} \times {\text{2) - 7)}} \\\ {\text{ = 2(}}{{\text{2}}^6} - 7) \\\ = {2^{6 + 1}} - 14 \\\ = {2^7} - 14 \\\$
Total number of ways of selecting at least 2 persons such that at least 1 boy and 1 girl included
$= {2^{12}} - 12{C_0}{ - ^{12}}{C_1} - ({2^7} - 14) \\\ = {2^{12}} - 1 - 12 - {2^7} + 14 \\\ = {2^{12}} - {2^7} + 1 \\\$

So, the correct answer is “Option D”.

Note: In order to solve such problems involving the number of ways of selection of items, the method of permutation and combination is the easiest to approach. In order to find the number of selections, use combinations and to find the number of different arrangements, use the method of permutation.