Let S denote the set of 4-digit numbers abcd such that a b c d, and P denote the set…

Question

Let $S$ denote the set of 4-digit numbers $\overline{abcd}$ such that $a > b > c > d$, and $P$ denote the set of 5-digit numbers having product of digits equal to 20. Then $n(S) + n(P)$ is equal to \_\_\_\_\_\_.

Answer: 260

Accepted Answer

For $S$: Any 4 distinct digits chosen from $\{0,1,\ldots,9\}$ form exactly one number with digits in strictly decreasing order. So $n(S) = \binom{10}{4} = 210$.

For $P$: 5-digit numbers whose digits multiply to 20. Possible digit multisets (using digits 1--9):

- $\{5,4,1,1,1\}$: arrangements $= \dfrac{5!}{3!} = 20$
- $\{5,2,2,1,1\}$: arrangements $= \dfrac{5!}{2!2!} = 30$

$n(P) = 20 + 30 = 50$

$n(S) + n(P) = 210 + 50 = 260$

Solutions · 1