I don't know how to do it, I have done it in two ways and getting different answers, can any one help with the right solution and explanation.

first method:
there are 3 boys and 7 girls, we have to make pairs with boys not together.

first boy goes with any of the 7 girls
second boy can go with any of the 6 girls
third boy can go with any of the 5 girls

now there are 4 girls left

so out of these one can go with any of the 3 left

at last there are 2 left to be paired

in all we have 7x6x5x3= 630

Second method:

10C2 - 3C2 = 42

I am more inclined towards this

Mom4MBA wrote:I don't know how to do it, I have done it in two ways and getting different answers, can any one help with the right solution and explanation.

first method:
there are 3 boys and 7 girls, we have to make pairs with boys not together.

first boy goes with any of the 7 girls
second boy can go with any of the 6 girls
third boy can go with any of the 5 girls

now there are 4 girls left

so out of these one can go with any of the 3 left

at last there are 2 left to be paired

in all we have 7x6x5x3= 630

Second method:

10C2 - 3C2 = 42

I am more inclined towards this

To make a valid pair a boy must be paired with a girl. First
choose a girl partner for each of the three boys.
The number of ways choosing 3 girls is = 7C3
The number of ways the chosen three girls can be paired
with three boys is = 3! × 7C3
The number of ways in which the remaining four girls can be
paired among themselves
4C2/2!
Hence, total number of ways= (3! × 7C3) × 3= 630


So Mom4MBA you were correct in ur 1st approach .

Thanks Shashank

There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

tienvunguyen wrote:There is a piece in this problem I don't understand. How can you have 630 pairs when the maximum of pairs taken out of 10 people is 10C2=45?

I guess your post just got duplicated over here.


As for the question,yes 10C2 =45 but what about the rest of the 8 people.They can also be paired up differently.

So u will get like 10C2 x 8C2 x 6C2 x 4C2 x 2C2 but over here I have also considered the case when 2 boys are paired up together.

I guess you left out the pairing of the other 8 ppl and hence got only 10C2.
Hope this helps you to understand the question carefully.