Combinations
- In mathematics, combination is the method from which we can calculate the possible arrangements from the given objects and where order of selection is not considered.
- If we have selected r number of things from the total n number of things then it can be denoted by nCr and given as,
nCr = n! / r! (n – r)!
This is the general formula for the combinations.
For example:
- We have 10 glass balls of different colors and we have to select any 5 of them, then the number of arrangements of selecting 5 glass balls from 10 glass balls is given by the combination as below,
- We have to select 5 glass balls from 10 glass balls.
- Hence, here n = 10 and r = 5
- Thus, the number of arrangements of selecting 5 glass balls from 10 glass balls is given by,
nCr = n! / r! (n – r)!
10C5 = 10! / 5! (10 – 5)!
10C5 = 3628800/ 120*120 = 252 ways
Thus, the number of ways of selecting 5 glass balls from 10 glass balls area 252 ways.
Some important tricks:
- If we have n objects and also, we have to select all the n objects then the number of possible ways of selection will be given by,
nCn= n! / n! (n – n)! = n! / n! 0!
We know that, 0! = 1
Hence, nCn= n! / n! (n – n)! = n! / n! 0! = n! / n! = 1
Thus, nCn = 1
- If we have n objects and out of which we have to select 0 objects then the number of possible arrangements is given by,
Here, n = n and r = 0
nC0 =n! / 0! (n – 0)! = n! / n! = 1
Thus, nC0 = 1
- If we have n number of objects and out of which we have to select the 1 object only then the number of possible ways of selections are given by,
Here, n = n, r = 1
nC1 = n! / 1! (n – 1)!
We know that, 1! = 1 and n! = n (n – 1)!
Hence, nC1 = n! / 1! (n – 1)! = n (n – 1)! / (n – 1)! = n
Thus, nC1 = n
Combinations are also of two kinds one includes repetition and other one includes no repetition.
Solved example:
1.) We have to make the committee of 3 boys and 2 girls from the 6 boys and 4 girls.
Then what will be the number of possible ways.
Ans:
- Here, we have to form the committee of 3 boys and 2 girls.
- So we have to select 3 boys from 6 boys and 2 girls from 4 girls.
- Hence in the form of combinations we can write as,
- 6C3: indicates selection of 3 boys from 6 boys
- 4C2: indicates selection of 2 girls from 4 girls.
6C3 = 6! / 3! (6 – 3)! = 6! / 3! 3! = 720 / 36 = 20
And 4C2 = 4! / 2! (4 – 2)! = 24/ 4 = 6
- Now, here committee is of boys and girls, if and comes then we have to take multiplication of combination.
- The number of ways of selecting 3 boys and 2 girls from 6 boys and 4 girls aregiven by,
6C3 *4C2 = 20*6 = 120
Thus, there are 120 ways of forming the committee of 3 boys and 2 girls from 6 boys and 4 girls.