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 6 of 10.

Ash said:   1 decade ago
Why nCr formula is not used here (6C2 x 4C2) , (6C3 x 4C1).?

Can any one explain me this

Jigs said:   1 decade ago
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.

Jyoti Nagpal said:   1 decade ago
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.

Kaushik said:   1 decade ago
Thank you Aditya..!

Djh said:   1 decade ago
@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.

Msrinivasarao said:   1 decade ago
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

Geet said:   1 decade ago
How to use ncr=ncn-r formula?why it is not used in second number 6c2 x 4c2?

Kanchan said:   1 decade ago
4C4
How we get value by it? Means what it is? What is the meaning of it?

Sam said:   1 decade ago
Is there anyone to explain why this particular question cannot be solved in the same method which were solved above?

Joe said:   1 decade ago
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?


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