Aptitude - Permutation and Combination - Discussion

Another easy method here;

It's given at least 1 boy meaning (>=1) so we can subtract calculate 0 boys selected and subtract it from the total combinations possible.
Combination(boys>=1) =total combination - combination (0boys selected),
= 10C4 - 6C0 * 4C4.
= 210 -1.
= 209.

1 case is missed out i.e., 4 girls can also be one more case.

So that makes up to 5 possible scenarios.

Let's break down the calculation step by step:

Given that there are 6 boys and 4 girls, we need to select 4 children such that at least one boy should be there.

We have four possible scenarios:
1. Selecting 1 boy and 3 girls.
2. Selecting 2 boys and 2 girls.
3. Selecting 3 boys and 1 girl.
4. Selecting 4 boys.

Now, let's calculate the number of ways for each scenario:

1. Selecting 1 boy and 3 girls:

- Number of ways to select 1 boy from 6 boys: 6C1 = 6 (combinations).
- Number of ways to select 3 girls from 4 girls: 4C3 = 4 (combinations).
- Total ways for this scenario: 6C1 x 4C3 = 6 x 4 = 24.

2. Selecting 2 boys and 2 girls:
- Number of ways to select 2 boys from 6 boys: 6C2 = 15 (combinations),
- Number of ways to select 2 girls from 4 girls: 4C2 = 6 (combinations),
- Total ways for this scenario: 6C2 x 4C2 = 15 x 6 = 90.

3. Selecting 3 boys and 1 girl:
- Number of ways to select 3 boys from 6 boys: 6C3 = 20 (combinations),
- Number of ways to select 1 girl from 4 girls: 4C1 = 4 (combinations),
- Total ways for this scenario: 6C3 x 4C1 = 20 x 4 = 80.

4. Selecting 4 boys:
- Number of ways to select 4 boys from 6 boys: 6C4 = 15 (combinations).

Now, add up the total ways for all four scenarios:

Total ways = (1) + (2) + (3) + (4) = 24 + 90 + 80 + 15 = 209.

So, there are 209 different ways to select four children from the group such that at least one boy is included.

A Group of 4 children, out of 10 children will be;
10c4 = 210.
As 4 girls are there;
So only 1 group is possible without the boy
Hence 210-1 = 209.

We may have(1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
In 1st step : ... + (6c1 x 4c3) +...6C4
In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?

Why don't the formula nCr = nC(n-r).
So 4C3 = 4C(4-3) = 4C1 apply to the other terms. Please Explain.

@All.

6C1 x 9C3.
Why can't be true?


For four Places we are taking conformally 1 from boys i.e 6C1 and there are total of 9(boys+girls) from we need to pick 3. So 9C3.

Hence the answer can be: 6C1 x 9C3.

Can anyone explain what's wrong with this?

Why can't we just simply do 6c1*9c3?!

In this, we've included that "minimum" condition of including that 1 boy and the rest 9 kids can be selected at random. why not? Someone, please explain.

@Huma. @Imnikesh.

4c3 becomes 4C1, Let me explain 4C3 means we have to take 3 girls out of 4 girls, generally by using the aforementioned formula was nCr = nC (n-r), this formula is only should be used on the condition that is value is half of the number greater than n value

i.e. nCr = 4C3, r=3 value which means are value is half of the value of n=4 increased (half of the value of n=4 is 2 I. E r=3 > 2) so that we applying nCr = nC (n-r) formula, 4C3 becomes 4C1. On the other case 6C1 we should not be applied this formula because r=1 value is smaller the half of the value of n=6 (half of the value of n=6 is 3 i.e r=1 < 3) that's why we could not be used aforesaid nCr formula.

Can anyone explain this 2nd step:-.

(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).

Why the nCr formula is applied to only (6C1 x 4C3) & (6C4) and not to all? Please explain.

Why can't we do 6C1*9C3?

First choose a boy from group of 6 and then choose any three from remaining 9 people.