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Day Care Problem

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Junior | Next Rank: 30 Posts
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by adi » Sat Mar 08, 2008 2:36 pm
There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A) 100
B) 90
C) 50
D) 45
E) 25

The answer should be D. Could anyone take me through this problem please?
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Re: Day Care Problem

by Stuart@KaplanGMAT » Sat Mar 08, 2008 3:43 pm
adi wrote:There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A) 100
B) 90
C) 50
D) 45
E) 25

The answer should be D. Could anyone take me through this problem please?
When we want to know the number of possible subgroups that we can make out of a big group, we use the combinations formula, which is:

nCk = n!/k!(n-k)!

n = total number of objects
k = number we want in each group

Here, we have 10 total objects and we're making groups of 2, so n=10 and k=2.

10C2 = 10!/2!(10-2)! = 10!/2!8! = 10*9*8!/2*8! = 10*9/2 = 90/2 = 45

If you're hoping to do well on the GMAT, the combinations formula is one that you need to know (and know well). The question you posted is on the low-end of difficulty for combinations questions (since it only requires knowing the basic formula and plugging in numbers from a simple scenario).
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by thinktank08 » Thu Apr 11, 2013 10:08 am
10C2 = (10*9)/2*1 = 45 :)

Keep it simple
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by Brent@GMATPrepNow » Thu Apr 11, 2013 10:18 am
thinktank08 wrote:10C2 = (10*9)/2*1 = 45 :)

Keep it simple
Exactly!

If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by vipulgoyal » Thu Apr 11, 2013 10:35 pm
let 10 students are
1,2---10

pairs of 1 =9
pairs of 2 = 8 bcz (1,2) and (2,1) are same
paire of 3 = 7
-
-
pair of 9 = 1= (9,10)
ans = 9+8+7+6+5+4+3+2+1 = 45
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Re: Day Care Problem

by Scott@TargetTestPrep » Mon Oct 05, 2020 6:44 am
adi wrote: ↑
Sat Mar 08, 2008 2:36 pm
 There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A) 100
B) 90
C) 50
D) 45
E) 25

The answer should be D. Could anyone take me through this problem please?
Solution:

Since 2 children are selected from 10 and the order of selection doesn’t matter, there are 10C2 = (10 x 9)/2 = 45 different pairs.

Answer: D

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