Question

In a group of 6 boys and 4 girls, four children are to be selected. In how many ways can they be selected such that at least one boy should be there
A) 109
B) 128
C) 138
D) 209

Answer 1

Hint: In this question it is given that, in a group there are 6 boys and 4 girls, we have to find in many ways four children are to be selected such that at least one boy should be there. So to find the solution we need to first find how many ways we can select 4 children from (6+5)=10 children( no restriction) and after that we will find the number of ways to form a group of 4 children where no boys are there. so from the total number of ways of selecting 4 children subtract none of the boys which gives us one boy or 2 boys or 3 boys or 4 boys, which is our required answer for at least one boy.

Complete step-by-step solution:
We have given 6 boys and 4 girls, intotal we have 6+4=10 children and we have to select 4 children from the given 10 children.
So by using combination we can say that we can select 4 children (randomly) from given 10 children in $${}^{10}C_{4}$$ =$$\dfrac{10!}{4!\left( 10-4\right) !}$$
     =$$\dfrac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 6!}$$
     =210.
Now we have to select 4 children where no boys are there, i.e, we can say that we have to select from four girls, so by using combination we can select 4 girls from 4 girls in $${}^{4}C_{4}$$ ways= 1 way,
Therefore, the total number of ways that we can select 4 children where no boys are there
= (no. of ways selecting 4 children randomly) - (no. of ways selecting 4 children where no boys are there)
= 210 - 1 = 209
Hence the correct option is option D.

Note: You can also solve this problem by another way, i.e, in this question it has been asked to find the number of ways of selecting 4 children where attest one boy should be there, that means you can select 1 boy or 2 boys or 3 boys or 4 boys, that implies that you will get 4 cases, and by solving you will ultimately get the same result, which is 209 way.