permutation: seven boys and 4 girls

Question

There are seven boys and four girls at a dance in how many ways can they form couples to dance if all the girls have partner?

Can someone teach me how to solve this?

Answer 1

We have

• 7 choices for the first girl
• 6 choices for the second girl
• 5 choices for the third girl
• 4 choices for the fourth girl

Now apply the Rule of Product to obtain

$$7\cdot 6 \cdot 5\cdot 4=840$$

(just as the previous problem)

Answer 2

So you have 4 girls. Assume the girls choose a partner each (one after the other), i.e. girl A chooses a partner first (she has 7 possible choices). Then girl B goes ahead and chooses one of the remaining 6 boys). Following are girls C and D.

Remember, that girl B has 6 choices left, no matter which of the boys girl A chose before here, so there are $7\times 6=42$ possible partner combinations of girl A and girl B with any of the boys.

This implies that in total you will have $7\times 6\times 5\times 4=840$ ways to form couples.