Question

The number of ways of choosing $3$ balls from $2$ yellow, $3$ green and $5$ blue balls such that at least $2$ blue balls are always selected.

Answer 1

Whenever we have to choose r different objects from n different objects, then we use the concepts of combination.
Make different cases according to questions and add their results to find the total number of combinations.

Formula used:
For selection of persons, we use combination (i.e., selection of n different objects taken r a ta a time) $^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Complete step by step solution:
Combination of n different objects taken $r$ at a time is given by: $^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
As given in the question we have to choose $3$ balls from $2$ yellow, $3$ green and $5$ blue balls such that at least $2$ blue balls are always selected .
So, we make different cases consisting $3$balls with at least $2$ blue balls are
$2$ blue balls and 1 yellow ball (out of $2$ yellow balls)
$2$ blue balls and $1$ green ball (out of $3$ green balls)
$3$ blue balls.

Case (I):
$2$ blue balls and 1 yellow ball (out of $2$ yellow balls) is $^{5}{{C}_{2}}$ $\times$ $^{2}{{C}_{1}}$(ways to select $2$ blue balls from $5$ is $^{5}{{C}_{2}}$=$\dfrac{5!}{2!\left( 5-2 \right)!}=\dfrac{5!}{2!\left( 3 \right)!}=5 \times 2=10$ ways and $1$ yellow ball from 2 is $^{2}{{C}_{1}}$=$\dfrac{2!}{1!\left( 2-1 \right)!}=2$ ways.) which is $10 \times 2=20$ ways.

Case (II):
$2$ blue balls and $1$ green ball (out of $3$ green balls) is $^{5}{{C}_{2}} \times ^{3}{{C}_{1}}$( ways to select $2$ blue balls is $^{5}{{C}_{2}}$=$\dfrac{5!}{2!\left( 5-2 \right)!}=\dfrac{5!}{2!\left( 3 \right)!}=5 \times 2=10$ways and $1$ green ball is $^{3}{{C}_{1}}$=$\dfrac{3!}{1!\left( 3-1 \right)!}=3$ways.) which is $10 \times 3$=$30$ways.

Case (III):
$3$ blue balls $^{5}{{C}_{3}}$(ways to select $3$ balls from 5 is $^{5}{{C}_{3}}$=$\dfrac{5!}{3!\left( 5-3 \right)!}=\dfrac{5!}{3!\left( 2 \right)!}=10$ ways).

These are the three possibilities, so we add these cases to get our result which is $20+30+10=60$ ways.

Note:
First identify that question is asking about the number of arrangements or combinations than according to the question and try to make possible cases then use the formula to find the case result and lastly add all the results to find the overall result.