Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 6)
6.
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
159
194
205
209
None of these
Answer: Option
Explanation:

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)
= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5
2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 + 90 + 80 + 15)
= 209.

Discussion:
92 comments Page 4 of 10.

Alex said:   1 decade ago
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?

Nikhil said:   1 decade ago
At least 1 boy = total selections - no boy.

= 10c4-(6c0*4c4).
= 210-1.
= 209.

Prabhu said:   1 decade ago
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).

Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?

Vijay said:   1 decade ago
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?

Hitesh said:   1 decade ago
What about when there was asked for at most 2 boys?

Aditi Muley said:   1 decade ago
In third step last 1 is 6C2, I am confused in the fourth step about how we solved 6C2 that we got 15 in result.

Shreu said:   1 decade ago
But the also there the children how it is possible?

Shilpa said:   1 decade ago
6c2= 6!/2!(6-2)!

= 6!/2!*4!
= 6*5*4! 2*4!

Then 4! will be cancel.
= 6*5/2.

By solving it.
= 15 got in result.

Maggi said:   1 decade ago
In 1st step :........+ (6c1 x 4c3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).

In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).

If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on.

Mayur said:   1 decade ago
One Question has many times came no one gave answer.

Why formula use for only first term ncr(n-r)?

Why not for other one, how can we know when we use this formula?


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