Question

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

Hint:

For strictly increasing or decreasing sequences of digits, the number of such integers is simply the number of ways to choose the digits ($\binom{n}{r}$).

Answer 1

Correct Answer: 179

First, find $n(S)$. The digits $a, b, c, d$ must be distinct elements from $\{0, 1, ..., 9\}$.
Since the order is strictly fixed ($a>b>c>d$), any combination of 4 distinct digits corresponds to exactly one valid number.
$n(S) = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
Next, find $n(P)$. We need 5-digit numbers with product of digits equal to 20.
The factors of 20 using single digits are combinations of $\{1, 2, 4, 5\}$.
Possible sets of 5 digits:
Case 1: $\{5, 4, 1, 1, 1\}$. Number of permutations = $\frac{5!}{3!} = 20$.
Case 2: $\{5, 2, 2, 1, 1\}$. Number of permutations = $\frac{5!}{2!2!} = 30$.
Note that 20 cannot be a digit, and 0 implies product 0.
$n(P) = 20 + 30 = 50$.
Total sum $= n(S) + n(P) = 210 + 50 = 260$.