Adance class consists of 22 students, of which 10 are women and 12 are men. if 5 men and 5 women are to be chosen and then paired off, how many results are possible?

solution:

I choose 5 women from a pool of 10 in 10C2 ways.

I choose 5 men from a pool of 12 in 12C2 ways.

So total number of ways of choosing in 10C2 x 12C2. Now I need to arrange them in 5 pairs. This is where I have a different solution. The solution says that there are 5! ways to arrange them in pairs.

But I cant seem to understand why? My reasoning is that for first pair position I need to choose 1 man from 5 and 1 woman from 5. So for the first position I have 5 x 5 choices (5 for man and 5 for woman). Similarly for the second position I have 4 x 4 choices and so on. Hence the total ways are 5! x 5!

So I calculate the total ways as 10C2 * 12C2 * 5! * 5!. Can anyone point the flaw in my reasoning for arranging the chosen men and women in pairs.

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation: