I'm trying to figure out why 6! gets multiplied by the number of choices for each man in diagram. I don't get it?
I don't understand the last part of this permutations/combinations question. Can someone explain?
0
$\begingroup$
algebra-precalculus
statistics
permutations
combinations
1 Answers
1
There are $6!$ elements in the set of arrangements of the women, and $7 \times 6\times 5$ elements in the set of arrangements of the men. And because the arrangements of the women and the arrangements of the men are totally independent of each other we take the Cartesian product of the two sets of arrangements to get the total number of ways of arranging the women and the men. And that means we multiply the number of elements in the two sets of arrangements, giving $6! \times 7 \times 6 \times 5$.