Problem Solving

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?

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).

10C2 = (10*9)/2*1 = 45

Keep it simple

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