Let S denote the set of 4-digit numbers abcd | vedclass

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(B) For $n(S)$,we need to choose $4$ distinct digits from the set ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$. Since the condition $a > b > c > d$ is given,once

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$260$

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(B) For $n(S)$,we need to choose $4$ distinct digits from the set ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$. Since the condition $a > b > c > d$ is given,once $4$ digits are chosen,there is only $1$ way to arrange them in descending order. Thus,$n(S) = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$. For $n(P)$,we need $5$-digit numbers whose product of digits is $20$. The prime factorization of $20$ is $2^2 	imes 5$. The possible sets of $5$ digits are: $1)$ ${5, 4, 1, 1, 1}$: The number of arrangements is $\frac{5!}{3!} = 20$. $2)$ ${5, 2, 2, 1, 1}$: The number of arrangements is $\frac{5!}{2!2!} = 30$. Thus,$n(P) = 20 + 30 = 50$. Therefore,$n(S) + n(P) = 210 + 50 = 260$.

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 the product of its digits equal to $20$. Then $n(S) + n(P)$ is equal to:

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