At a party there are 20 girls and 20 boys.
How many dance pairs can be formed if boys may dance with boys, and girls with girls?
I have taken a smaller set of 3 girls and 3 boys to try and understand this problem. Altogether this is 6 people. We can form pairs (where the order of the partners do not matter) in such a way:
- From 6 people, choose 2 of them to form a pair -> C(6,2) = 15, then
- from 4 people, choose 2 of them to form the next pair -> C(4,2) = 6, then
- from the last 2 people, choose 2 to form the final pair -> C(2,2) = 1
Therefore forming (15)(6)(1) = 90 dance pairs
Am I understanding this question correctly? Is my reasoning sound, if so can this be extended to the problem of 20 boys and 20 girls?
Any help is appreciated.
Best Answer
@JMoravitz gave you beautiful comment to teach it to you . I am writing this answer to give you another view.
It is said that boys can dace with boys , girls can dance with girls , boys and girls can dance ,as well . So, it is understod that every one can dance with anyone .Hence ,we have $40$ people to dance.
Now , i want you to think that these $40$ people are like $40$ distinct balls . Moreover , think that dancing is the same as getting in a box with another ball , and all boxes are same , because it is not mentioned that dancing is different type. Lets assume that all of them dacing tango.
Now , the question turned out to be placing $40$ distinct balls into $20$ identical boxes where each box will have exactly $2$ balls.
To solve it , firstly think that boxes are different , so there are $\binom{40}{2}$ ways for the first box , there are $\binom{38}{2}$ ways for the second box , .... , there are $\binom{2}{2}$ ways for the $20th$ box . Multiplicaiton of these binomal coefficients is equal to $\frac {40!}{2^{20}}$
However , the boxes were identical , to make them identical again divide the result by $20!$.
So , the answer is $$\frac {40!}{2^{20} \times 20!}$$