There are 12 children in a party. For a game they have to be paired up...
To determine the number of different pairs that can be made for the game, we can use the concept of combinations.
Combination is a way to select items from a larger set without considering the order of the items. In this case, we want to pair up the 12 children, so we need to find the number of combinations of 2 children that can be formed from a group of 12.
The formula to calculate the number of combinations is given by:
nCr = n! / (r!(n-r)!)
Where n is the total number of items, r is the number of items to be selected, and ! denotes factorial.
Now let's calculate the combinations:
n = 12 (total number of children)
r = 2 (number of children to be selected for each pair)
Using the formula:
12C2 = 12! / (2!(12-2)!)
= 12! / (2!10!)
= (12 * 11 * 10!) / (2! * 10!)
= (12 * 11) / 2!
= 132 / 2
= 66
Therefore, there are 66 different pairs that can be made for the game.
Hence, the correct answer is option D) 66.
There are 12 children in a party. For a game they have to be paired up...
For this question we could us traingular number formula as
person __ pair
1_0
2_1
3_3
4_6
by formula we have 12 person which form pairs equal to traingular no. of 11 i. e..... (n*(n+1) ) /2
(11*11+1) /2=66