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

Question

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

Answer: 260

Accepted Answer

for set $S$ : 4-digit numbers $abcd$ with $a>b>c>d$

digits are chosen from ${0,1,2,\ldots ,9}$

since order is strictly fixed, any 4 distinct digits chosen will form exactly one such number. number of ways to choose 4 digits from 10 is ${}^{10}{C}_{4}$

$n(S)=\frac{10	imes 9	imes 8	imes 7}{4	imes 3	imes 2	imes 1}=210$

for set $P$ : 5-digit numbers with product of digits $=20$

possible sets of digits (using digits 1-9):

case 1: ${5,4,1,1,1}$

number of arrangements $=\frac{5!}{3!}=20$

case 2: ${5,2,2,1,1}$

number of arrangements $=\frac{5!}{2!2!}=30$

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

calculating final value.

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

the value of $n(S)+n(P)$ is 260 .

Solutions · 1