Exercise 1621Combinatorics › Counting principlesL1Added 3 Nov 2026

The multinomial theorem

#multinomial#product rule#word counting

Statement

How many ways can $10$ distinct balls be distributed into $3$ boxes labeled $A,B,C$ , with exactly $4$ in $A$ , $3$ in $B$ , and $3$ in $C$ ?

Hint

Multinomial coefficient $\binom{10}{4,3,3}=\frac{10!}{4!3!3!}$ .

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Choose first. $\binom{10}{4}$ ways to pick $4$ balls for $A$ . Then $\binom{6}{3}$ for $B$ , and $1$ way for $C$ (remaining $3$ ).

Total. $\binom{10}{4}\binom{6}{3}=\frac{10!}{4!\,6!}\cdot\frac{6!}{3!\,3!}=\frac{10!}{4!\,3!\,3!}=4200$ .

Conclusion. There are $4200$ ways.