Combinatorics problem

This problem is from Richard A. Brualdi's Introductory Combinatorics. Though I've somehow derived the answer, the logic of my solution doesn't make much sense. Help me solve this.

Q. At a party there are 15 men and 20 women. How many ways are there to form 15 couples consisting of one man and one woman?

#Combinatorics

Easy Math Editor

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Comments

There are $\binom{20}5$ ways to choose the 5 women who are not in a couple and $15!$ ways to pair 15 men with 15 women so the answer is $\large\boxed{20274183401472000}$ .

Ryan Soedjak - 7 years, 7 months ago

Great.

Haris Bin Zahid - 7 years, 7 months ago

15C15 * 20C15 = 15504

Feroz Baig - 7 years, 7 months ago

One unique man and one unique woman can be chosen in 15504 ways. Since you've to form 15 couples, you need to multiply this answer by 15!

Haris Bin Zahid - 7 years, 7 months ago

All men end up in couples regardless of what happens.Now, Down to the women, there are 20C15 to choose women who are in couples.. Multiply that with 15! because there are 15 couples in total.

20C15=15504 15504 x 15!= 20274183401472000

Daud Shehzad - 7 years, 5 months ago

35C2

Josephite Tirtho - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$