Aptitude - Permutation and Combination - Discussion

@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.

Its easy.

10c4 - 4c4 =209.

What if the question was at least 2 boys? Need help.

@Udaya.

May I know how could you find out the answer from below calculation?

So the number of ways the selection can be made such that at least one boy should be there is =10C4 - 4C4 = 209.

Please explain this.

Should not we have to arrange all the children? In this we have just make combinations.

Can anyone explain how to derive this calculation? Please.

Why is it wrong 6C1 * 9C3?

Can it not be written as 6 * 9C3?

@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. :)

= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
(6C1 x 4C3) AND (6C4)applied to ncr.