Question

In a party there are $25$ men and $20$ women, then in how many ways a couple (one man with one woman) can be formed, when $2$ particular men and $5$ particular women refuse to be part of any couple?

• A. $500$
• B. $460$
• C. $345$
• D. $300$

Answer 1

The correct option is C $345$

Given : Men $=25$, women$=20$
Now,
Selection of one man (two men do not want to be coupled) out of $25$
$={}^{(25-2)}{C}_1={}^{23}{C}_1$ and
Selection of one woman (five women do not want to be coupled) out of $20$
$={}^{(20-5)}{C}_1={}^{15}{C}_1$
$\therefore$ Required number of couples formed $={}^{23}{C}_1\cdot {}^{15}{C}_1=23\times 15=345$