Question

Let \( S \) be the number of 4-digit numbers \( abcd \), where \[ a>b>c>d \] and let \( P \) be the number of 5-digit numbers \( abcde \), where the product of digits is 20. Find \( S + P \):

Hint:

For digit-product problems, always factor the number and list valid digit combinations carefully.

Answer 1

Correct Answer: 260

Step 1: Find the value of \( S \).
Digits are chosen from \( \{0,1,2,\ldots,9\} \) such that \[ a>b>c>d \] This is equivalent to choosing 4 distinct digits and arranging them in decreasing order. Number of ways: \[ S = \binom{10}{4} = 210 \]
Step 2: Find the value of \( P \).
We want 5-digit numbers whose digit product is 20. Prime factorization: \[ 20 = 2^2 \times 5 \] Possible digit combinations (excluding zero) are permutations of: \[ (1,1,4,5,1),\; (1,1,2,2,5) \] Counting all valid permutations gives: \[ P = 50 \]
Step 3: Compute the final value.
\[ S + P = 210 + 50 = 260 \]