enriqueta26 wrote:COULD SOMEONE HELP ME WITH THIS ONE, PLEASE?

THERE ARE 100 CHILDREN. A PAIR OF CHILDREN IS TO BE SELECTED TO PLAY. AT MOST, HOW MANY DIFFERENT PAIRS OF CH ARE POSSIBLE?

THANK YOU MUCH!!

first selection is to be done from 100 childres so firsy children can be selected in 100 ways
to choose second child we are left with 99 student
so total number of different pair of child 100*99=9900

From Enriqueta's post, it does not seem like the order of selection matters. Picking John and Kelly is the same as picking Kelly and John. For 2 pairs to be different, they must not include the same 2 children. This matters

Amising's solution would be correct only if order of selection mattered.

For instance, suppose there are 3 children and we want to figure out how many pairs are possible. Amising's solution would be:
>> choose the first child from 3 options.
>> choose the second child from 2 options.
>> a total of 6 pairs.

Actually there are fewer than 6 pairs. With 3 children A, B, and C the only possibilities are (A,B), (A,C) and (B,C). This is because (A,B) and (B,A) is the same pair. The solution with 3 children to pick from is 3, not 6.

This is a straightforward combination problem. How many ways can we group 100 kids into pairs is 100C2 (choose 2 from 100, order is irrelevan). This is 100!/(2!98!) = 4,950 pairs possible

-Patrick

thank you guys but this problem doesn look it a combination one!

We have 100 children and he have to choose a different pairs to play. At most, we supossed to have 45 different pairs possible to choose following the answer.??

What are the answer choices Enriqueta? and what is the source of the question please?

I can guarantee you that there are more than 45 pairs possible if you have 100 children to pick from. My understanding of the question is that there are 100 kids and we want to know how many ways can we select a group of 2 kids (a pair). Maybe I misunderstood the question?

-Patrick

Thank you for your attention Patrick!!

My source is Gmat practice exam. what I had written is:

There are 100 children, and a pair of children is to be selected to play.

At most, how many different pairs of children are possible?

I really don't know why the answer is supposed to be 45

There is a small chance I missed some information

Thank you!

Logically, this is why there must be more than 45 pairs possible if we have 100 children to pick from.

Suppose the children are numbered C1 through C100.

We could have pairs {C1,C51}, {C2,C52}, {C3,C53}...{C50,C100}. This alone already gives us 50 possible pairs. But there are many more pairs still. For example, instead of pairing C1 with C51, we could have {C1,C2}... At the end of the day, there are a lot more than just 45 possible pairs.

If we had 10 children to pick from (instead of 100), then the answer would be 45.

All of this assumes that I'm not misunderstanding the problem still. If I am, then all bets are off. Do you still have access to the question? If so please proofread and post the answer choices as well.

-Patrick

Patrick,

I do not have access to the question now, but I will try to recheck the question again.

Many thanks!!