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 8 of 10.

Shiva said:   8 years ago
I am confused in the second step, Please explain it.

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

It's one of the formulae.

Heenu said:   7 years ago
Easiest way:

Since 4 students are to be selected and there are 4 girls, hence we don't want a combination consisting of all girls: 4c4 =1.
Total no of combinations available: (Total students= 6 boys + 4 girls =10 and 4 need to be selected) 10c4= 210.
Hence favorable combinations = 210 -1 = 209.

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?

Tab said:   7 years ago
Thanks @Udaya.

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

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.

Shivam said:   6 years ago
No boys selected=4c4 ways i.e 1.

Selection of 4 from all =10c4 ways=10*9*8*7/4*3*2*1=210 ways.
Atleast 1 boy=210-1 = 209 ways.
(3)

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

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


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