Aptitude - Permutation and Combination - Discussion

Another easy method is

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

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?

Hey Thiyanesh you are little bit confused, read the question again.

In second step

Why we use the nCr formula only for 4C3 and 6C4?

It can also be done in another way... we can calculate total no of ways to select the children..ie 6+4=10 10C4 and subtract ways in which no boy is there ie all girls 4C4...
10C4-4C4= 209
Calculation would be easy in this way

It is very simple. Only read carefully after that ans we get.

Udaya its easy. But its more calculative.

Udaya, your answer was really simpler.

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 :)

Hi it is simple, read question ones again

We have 6-boys and 4-girls we have to select only 4 members from the groups so in that once condition least 1 boy means one boy or more 1 than boy or all the boys only

(6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)=209