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?
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 numberof ways |
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
|
||||||||||||||||||||
| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. |
Discussion:
92 comments Page 7 of 10.
Hmy nj said:
9 years ago
Why is it wrong 6C1 * 9C3?
Alok said:
9 years ago
Can it not be written as 6 * 9C3?
Divyae said:
9 years ago
@Alok and @Hmy Nj
We cannot do 6C1 * 9C3 as there will be some repeated selections in this case.
Eg, B1, B2, B3, B4, B5, B6 G1, G2, G3, G4
Consider in the first step you selected B1, you have options of selecting any boy in step 2.
Now, if you consider selecting B2 in the first step, then you can select B1 in step 2. Thus it creates similar selections.
I hope it is clear. :)
We cannot do 6C1 * 9C3 as there will be some repeated selections in this case.
Eg, B1, B2, B3, B4, B5, B6 G1, G2, G3, G4
Consider in the first step you selected B1, you have options of selecting any boy in step 2.
Now, if you consider selecting B2 in the first step, then you can select B1 in step 2. Thus it creates similar selections.
I hope it is clear. :)
Badri said:
9 years ago
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
(6C1 x 4C3) AND (6C4)applied to ncr.
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
(6C1 x 4C3) AND (6C4)applied to ncr.
DANIEL said:
9 years ago
Why not 4 girls?
Neema said:
9 years ago
Not mention that how many person should we take at a time, the how it will be solved?
Shivaraj said:
9 years ago
@Neema.
It's mentioned, read question clearly.
It's mentioned, read question clearly.
Pranali said:
9 years ago
In first step (6c2 * 4c2).
Send step ( (6*5/2*1) * (4*3/2*1) ).
How?
I can't understand it
Send step ( (6*5/2*1) * (4*3/2*1) ).
How?
I can't understand it
Harika said:
8 years ago
I'm bit confused that why here use multiplication? Please tell me.
Tarun said:
8 years ago
@Harika.
It's because those two events are needed to complete the work. If any one event is needed to complete the work, we use addition. Hope this helped.
It's because those two events are needed to complete the work. If any one event is needed to complete the work, we use addition. Hope this helped.
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