Question

A box contains $5$ different red and $6$ different white balls. In how many ways can $6$ balls be selected so that there are at least two balls of each colour?

Answer 1

The selection of $6$ balls, consisting of at least two balls of each colour from $5$ red and $6$ white balls, can be made in the following ways :

$\left(i\right)$By selecting $2$ red balls out of $5$ and $4$ white balls out of $6$. This can be done in ${}^5{C}_2{\times }^6{C}_4$ ways.

$\left(ii\right)$By selecting $3$ red balls out of $5$ and $3$ white balls out of $6$. This can be done in ${}^5{C}_3{\times }^6{C}_3$ ways.

$\left(iii\right)$By selecting $4$ red balls out of $5$ and $2$ white balls out of $6$. This can be done in ${}^5{C}_4{\times }^6{C}_2$ ways.

Since the selection of $6$ balls can be completed in any one of the above ways. Hence, by the fundamental principle of addition, the total number of ways to select the balls

${=}^5{C}_2{\times }^6{C}_4{+}^5{C}_3{\times }^6{C}_3{+}^5{C}_4{\times }^6{C}_2$

$=10\times 15+10\times 20+5\times 15=425$