Question

Out of 6 boys and 4 girls, a group of 7 is to be formed. In how many ways can this be done, if the group is to have a majority of boys?
A. 120
B. 80
C. 90
D. 100

Answer 1

Hint: We first separate the groups in which the boys have the majority. We separately find the number of ways we can arrange people and add them to find the final number of possible ways.

Complete step-by-step answer:
There are in total 6 boys and 4 girls out of which a group of 7 is to be formed
The given condition is to have a majority of boys in the group.
So, the number of possible combinations for group of 7 will be with
(i) 4 boys and 3 girls (ii) 5 boys and 2 girls (iii) 6 boys and 1 girls
We separately find the number of ways. We take the first situation where only one attends.
The choice for the first condition is for 4 boys and 3 girls out of 6 boys and 4 girls.
The number of choices will be $ {}^{6}{{C}_{4}}\times {}^{4}{{C}_{3}}=\dfrac{6!}{4!\times 2!}\times \dfrac{4!}{1!\times 3!}=15\times 4=60 $ ways.
The choice for the second condition is for 5 boys and 2 girls out of 6 boys and 4 girls.
The number of choices will be $ {}^{6}{{C}_{5}}\times {}^{4}{{C}_{2}}=\dfrac{6!}{5!\times 1!}\times \dfrac{4!}{2!\times 2!}=6\times 6=36 $ ways.
The choice for the third condition is for 6 boys and 1 girl out of 6 boys and 4 girls.
The number of choices will be $ {}^{6}{{C}_{6}}\times {}^{4}{{C}_{1}}=\dfrac{6!}{6!\times 0!}\times \dfrac{4!}{1!\times 3!}=1\times 4=4 $ ways.
Total will be $ 60+36+4=100 $ . The correct option is D.
So, the correct answer is “Option D”.

Note: We also could have separated the first event in person based but the number of ways is same for both of them and that’s why we multiplied 2 with the value to get the total. There are some constraints in the form of $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!} $ . The general conditions are $ n\ge r\ge 0;n\ne 0 $ .