Question

Among set of 5 green balls and 3 blue balls how many selection of 5 balls can be made such that atlest 3 of them are green balls

Answer 1

8c5-(5c2+3c3)=56-11=45.
here we select (5 balls from 8 balls ) =8c5
then subtracting cases that has less than 3 green ball cases ,,,that leaves with only one case i.e 5c2+3c3 =>here we don't take 5c1 because we have only 3 blue balls.

Answer 2

Answer:

46 selection of 5 balls can be made such that at least 3 of them are green balls

Step-by-step explanation:

Green balls = 5

Blue balls = 3

Now we are supposed to find how many selection of 5 balls can be made such that at least 3 of them are green balls .

So, No. of ways of election of 5 balls can be made such that 3 of them are green balls = $^5C_3 \times ^3C_2$

No. of ways of election of 5 balls can be made such that 4 of them are green balls = $^5C_4 \times ^3C_1$

No. of ways of election of 5 balls can be made such that 5 of them are green balls = $^5C_5$

So, Total no. of ways = $^5C_3 \times ^3C_2+^5C_4 \times ^3C_1+^5C_5$

Total no. of ways = $\frac{5!}{3!2!} \times\frac{3!}{2!}+\frac{5!}{4!} \times \frac{3!}{2!}+\frac{5!}{5!0!}$

Total no. of ways = $46$

Hence 46 selection of 5 balls can be made such that at least 3 of them are green balls