Given 5 different green toys, 4 different blue toys and 3 different red toys, how many combinations of toys…
Given 5 different green toys, 4 different blue toys and 3 different red toys, how many combinations of toys can be chosen taking at least one green and one blue toy?
$32 \times 16 \times 4$
$31 \times 15 \times 4$
$32 \times 16 \times 8$
$31 \times 15 \times 8$
Solution
Selecting atleast 1 green toy out of 5 can be done in ${ }^5 \mathrm{C}_1+{ }^5 \mathrm{C}_2+{ }^5 \mathrm{C}_3+{ }^5 \mathrm{C}_4+{ }^5 \mathrm{C}_5+=31$ ways.
Selecting atleast 1 blue toy out of 4 can be done in ${ }^4 \mathrm{C}_1+{ }^4 \mathrm{C}_2+{ }^4 \mathrm{C}_3+{ }^4 \mathrm{C}_4=15$ ways.
Selecting red toys without restriction can be done in ${ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3=8$ ways.
Total number of ways $=31 \times 15 \times 8$