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.

Gayathri m said:   1 week ago
The second step is not determined.

i.e (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).

Nithin said:   5 years ago
Why can't we do 6C1*9C3?

First choose a boy from group of 6 and then choose any three from remaining 9 people.

Sathvika said:   6 years ago
4c2 = 4!/2!*(4-2)!.
= 4 * 3/2 * 2,
= 3.

Dhanraj said:   6 years ago
Thanks for explaining it in detail @Shivam.

Aniket said:   7 years ago
@All.

6c4 = 6c2.

Because 6c4= 6!/(4!*2!) and 6c2=6!/(2!*4!) both are same that's why 6c4=6c2.

Imnikesh said:   7 years ago
How 6c4 is reduced to 6c2?

Tab said:   7 years ago
Thanks @Udaya.

Sandy said:   7 years ago
@ALL.

So you can select 1 boy in 6C1 ways. So the rest 3 of them can be any gender right. So why can't it be done in 9C3 ways?

Why is the final answer not 6C1*9C3?

Neema said:   9 years ago
Not mention that how many person should we take at a time, the how it will be solved?

Aarav said:   7 years ago
We can also apply nCr = n!/((r!)(n-r)!).

It's one of the formulae.


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