Question

A father buys nine different toys for his four children. In how many ways can he give one child three toys and the remaining three children two toys each? 5. A party of nine people consists of five men and four women, and a group of four people is to be chosen at random from this party. In a how many ways can a group of four be chosen that contains at least three women?

Answer 1

Solution:

The one child who will be given 3 toys can be chosen in $\left(\begin{array}{l}4 \\ 1\end{array}\right)$ ways.

His/her 3 toys can be chosen in $\left(\begin{array}{l}9 \\ 3\end{array}\right)$ ways.

That leaves 6 toys to be distributed among 3 kids, 2 to each.

Line up these three kids in order of age.

The toys for the oldest can be chosen in $\left(\begin{array}{c}6 \\ 2\end{array}\right)$ ways, and for each way the toys for the next oldest can be chosen in $\left(\begin{array}{c}4 \\ 2\end{array}\right)$ ways, for a total of $\left(\begin{array}{c}4 \\ 1\end{array}\right)\left(\begin{array}{c}9 \\ 3\end{array}\right)\left(\begin{array}{l}6 \\ 2\end{array}\right)\left(\begin{array}{l}4 \\ 2\end{array}\right)$

$=4\times84\times15\times6=30,240$

(5):

Out of 5 men, 4 women, a group of 4 is to be made with at least three women.

Total required ways are $\left(\begin{array}{l}4 \\ 3\end{array}\right)\left(\begin{array}{l}5 \\ 1\end{array}\right)+\left(\begin{array}{l}4 \\ 4\end{array}\right)\left(\begin{array}{l}5 \\ 0\end{array}\right)$

$=4\times5+1\times1=21$