Question ID: #847
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…………
Solution:
**Calculate n(S):**
We need to form 4-digit numbers with distinct digits in strictly descending order ($a > b > c > d$).
Any selection of 4 distinct digits from $\{0, 1, 2, …, 9\}$ can be arranged in exactly one way to satisfy the condition.
Note that since $a > b > c > d$, the largest digit $a$ can never be 0, so the 4-digit constraint is automatically satisfied.
$$ n(S) = ^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 $$
**Calculate n(P):**
Product of 5 digits = 20.
Prime factors of 20 = $2^2 \times 5$.
Possible combinations of 5 digits (filling the rest with 1s):
Case 1: Digits are $\{4, 5, 1, 1, 1\}$.
Number of permutations = $\frac{5!}{3!} = \frac{120}{6} = 20$.
Case 2: Digits are $\{2, 2, 5, 1, 1\}$.
Number of permutations = $\frac{5!}{2!2!} = \frac{120}{4} = 30$.
Total $n(P) = 20 + 30 = 50$.
**Final Result:**
$$ n(S) + n(P) = 210 + 50 = 260 $$
Ans. (260)
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