Aptitude - Permutation and Combination - Discussion

1 boy can be selected in 6 ways.

The rest 3 can be selected from the remaining 9, in 9C3 ways=84 ways.

Hence the total no: of ways = 6*84 = 504.

Is it right?

Another easy method is

1-(4C4/10C4)
=1-(1/210)
=209/210
Therefore 209 times out of possible 210.

4C4
How we get value by it? Means what it is? What is the meaning of it?

How to use ncr=ncn-r formula?why it is not used in second number 6c2 x 4c2?

We have 6-boys and 4-girls we have to select only 4 members from the group In that at least one boy should be there.so
we cab write
total number of ways-no boy can be selected

10c4-4c4=210-1
=209

@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.

Thank you Aditya..!

Another Shortcut Method: Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 childern so, 4c4 so selecting atleast one boy 10c4-4c4=210-1=209.

Another Shortcut Method:.

Total student=6+4=10 The number of ways to selecting 4 girls=10c4, The number of ways to selecting no boys= 4c4 because total boys are=6 so 10-6=4 to select 4 children so, 4c4 so selecting at-least one boy 10c4-4c4=210-1=209.

Why nCr formula is not used here (6C2 x 4C2) , (6C3 x 4C1).?

Can any one explain me this