Question

In a group of 6 boys and 5 girls, 5 students have to be selected. In how many ways it can be done so that at least 2 boys are included

Answer 1

Answer:

431 ways

Step-by-step explanation:

Given,

Number of boys = 6,

Number of girls = 5,

If 5 students are selected such that there must be at least 2 boys,

Then, total possible ways = 2 boys and 3 girls + 3 boys 2 girls + 4 boys 1 girl + 5 boys,

$=^6C_2\times ^5C_3 + ^6C_3\times ^5C_2+^6C_2\times ^5C_1+^6C_5\times ^5C_0$

$=\frac{6!}{4!2!}\times \frac{5!}{3!2!}+\frac{6!}{3!3!}\times \frac{5!}{3!2!}+\frac{6!}{4!2!}\times \frac{5!}{4!1!}+\frac{6!}{5!1!}$

= 15 × 10 + 20 × 10 + 15 × 5 + 6

= 150 + 200 + 75 + 6

= 431