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 10 of 10.
Mohit said:
1 decade ago
Can anyone please help me to reduce the time spent in lengthy calculations. I understood how it can be solved. But calculations take so much time and we don't have much time in solving one particular question.
If I try to solve fast, small silly mistake happens and could not get answer in 1st attempt which increases the time spent to re-check it.
If I try to solve fast, small silly mistake happens and could not get answer in 1st attempt which increases the time spent to re-check it.
Imran said:
1 decade ago
See 4C4 means equal to 1 bcoz, there is only one possible way to have all 4 boys i.e 4 out of 4,
So, 10C4 - 4C4 now apply on formula N!/n!(N-n)!.
10!/4!(10-4)! -4C4 = 10!/4!(6!)-4C4.
= 10*9*8*7*6*5*4*3*2*1/4*3*2*1 * 6*5*4*3*2*1 -4C4.
= 10*9*8*7/4*3*2*1 -4C4 (bcoz 6*5*4*3*2*1 cut up and down).
= 5040/24 -4C4.
= 210 -4C4.
= 210 - 1 = 209 answer, (bcoz 4C4 = 1).
So, 10C4 - 4C4 now apply on formula N!/n!(N-n)!.
10!/4!(10-4)! -4C4 = 10!/4!(6!)-4C4.
= 10*9*8*7*6*5*4*3*2*1/4*3*2*1 * 6*5*4*3*2*1 -4C4.
= 10*9*8*7/4*3*2*1 -4C4 (bcoz 6*5*4*3*2*1 cut up and down).
= 5040/24 -4C4.
= 210 -4C4.
= 210 - 1 = 209 answer, (bcoz 4C4 = 1).
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