# Answer to Question #11259 in Statistics and Probability for Ke_091

Question #11259

In a group of 15 boys, there are 6 scouts. In how many ways can 8 boys be selected, so as to include

(i) exactly 3 scouts

(ii) at least 3 scouts

(iii) at the most 3 scouts

(i) exactly 3 scouts

(ii) at least 3 scouts

(iii) at the most 3 scouts

Expert's answer

i) So we pick 3 scouts among 6: 6!/(3!3!)=20 and other 3 boys among non-scouts 9!/(6!3!)=84. Number of ways we're asked about - 20*84=1680.

ii) At least 3 scouts, so it's fine when ther is 3,4,5 or 6 scouts. From previous we know that N(3)=1680. Now analogically N(4)=6!/(4!2!)*9!/(7!2!)=540,

N(5)=6!/(5!1!)*9!/(8!1!)=54 and N(6)=1. So the number we need is

N(=>3)=N(3)+N(4)+N(5)+N(6)=2275.

iii) Total number of variants to pick 6 boys among 15 is 15!/(9!6!)=5005. Now using this number:

N(<=3)=5005-N(=>3)+N(3)=4410.

ii) At least 3 scouts, so it's fine when ther is 3,4,5 or 6 scouts. From previous we know that N(3)=1680. Now analogically N(4)=6!/(4!2!)*9!/(7!2!)=540,

N(5)=6!/(5!1!)*9!/(8!1!)=54 and N(6)=1. So the number we need is

N(=>3)=N(3)+N(4)+N(5)+N(6)=2275.

iii) Total number of variants to pick 6 boys among 15 is 15!/(9!6!)=5005. Now using this number:

N(<=3)=5005-N(=>3)+N(3)=4410.

## Comments

## Leave a comment