Aptitude - Permutation and Combination - Discussion

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.

I completely don't get it! Combintations:

1B+3G
2B+2G
3B+1G
4B

Those are very clear, no doubts.

so, the final answer is (1b+3g)+(3b+2g)+(3b+2g)+4b

Let's see the first one.

1b+3g=BGGG (as we have 4 spots). So, for the boy we have 6 different options (as any of the 6 boys can be chosen). We have 3 spots for girls and 4 girls to fill those 3 spots. So, for the first of the 3 girl spots there are 4 different options to choose from (as we have 4 girls), for the second spot we have only 3 girls left (as the forth girl already took the first spot), and for the 3d spot we have 2 girls left. So, the total result for 1b+3g would be 6x4x3x2+ we have to account the other combinations (2b+2g, 3b+1g, 4b). So, how the result is only 209?

@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.

This question can be solved in a more simpler way .
The requirement is that there should be atleast one boy out of the 4 children selected.

So we get the output by subtracting
the number of ways in which no boy is selected from
the number of ways in which 4 children can be selected from total(i.e,10)

Here goes the solution :
Total no. of children(including boys n girls) = 10
The number of ways in which 4 children can be selected from 10 children is = 10C4
The no. of ways in which no boy is selected = 4C4
So the number of ways the selection can be made such that atleast one boy should be there is =
10C4-4C4=209.

As simple as that ...Rite :)

@joe

Your method is wrong because you cannot multiply 6C1*9C3 as it is a basic contradiction of m*n rule.

The multiplication of m*n is possible only if they are the respective no of possibilities of two separate stages for the outcome of an event.

Here the first stage is boys(6c1) and the second stage will be both boys+girls(9C3).

Hence m*n cannot be done here as the second event is not a separate(girls only)stage. Hope I put it across clearly.

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?

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.

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.

See 4C4 means equal to 1 bcoz, there is only one possible way to have all 4 boys i.e 4 out of 4,

So, 10C4 - 4C4 now apply on formula N!/n!(N-n)!.

10!/4!(10-4)! -4C4 = 10!/4!(6!)-4C4.

= 10*9*8*7*6*5*4*3*2*1/4*3*2*1 * 6*5*4*3*2*1 -4C4.

= 10*9*8*7/4*3*2*1 -4C4 (bcoz 6*5*4*3*2*1 cut up and down).

= 5040/24 -4C4.

= 210 -4C4.

= 210 - 1 = 209 answer, (bcoz 4C4 = 1).

@Alok and @Hmy Nj

We cannot do 6C1 * 9C3 as there will be some repeated selections in this case.

Eg, B1, B2, B3, B4, B5, B6 G1, G2, G3, G4

Consider in the first step you selected B1, you have options of selecting any boy in step 2.
Now, if you consider selecting B2 in the first step, then you can select B1 in step 2. Thus it creates similar selections.

I hope it is clear. :)