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 3 of 10.
Shanmuga kumar said:
1 decade ago
How do you get 4 members in group? please explain.
Kashika shekhar said:
1 decade ago
I am not able to apply different types of logic in different questions. Please tell me some logic for appying the formulas.
Anusha said:
1 decade ago
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).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
Why to change only the first term and last term with formula. Why don't the formula apply to the other terms. Please Explain?
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).
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?
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.
Baba said:
1 decade ago
You could also take a shortcut and say that there are 10*9*8*7 / 4! = 210 ways to arrange 10 people in a group of 4. Then subtract the number of ways to arrange 4 girls in a group of 4 4*3*2*1 / 4! = 1. So 210 Total - 1 All Girl = 209 At least 1 Boy.
Kiran said:
1 decade ago
There is a trick here just oppose the question like following:
We have 6 Boys and 4 Girls,
1) Total No. of Possible ways = 10c4 -->210 (selecting 4 out of 10 Children).
2) Tricky here(Oppose) select no boy at all i.e select 4 children from only girls = 4C4 --> 1.
Finally : Subtract 2nd from 1st 210-1 = 209 Answer.
We have 6 Boys and 4 Girls,
1) Total No. of Possible ways = 10c4 -->210 (selecting 4 out of 10 Children).
2) Tricky here(Oppose) select no boy at all i.e select 4 children from only girls = 4C4 --> 1.
Finally : Subtract 2nd from 1st 210-1 = 209 Answer.
Nishant Khanorkar said:
1 decade ago
Best way to solve such problem is by going the other way round.
At least 1 boy = total - all girls.
So,
10c4-4c4
= (10*9*8*7)/(1*2*3*4) - 1.
= 210 - 1.
= 209.
And here it is. out required answer.
At least 1 boy = total - all girls.
So,
10c4-4c4
= (10*9*8*7)/(1*2*3*4) - 1.
= 210 - 1.
= 209.
And here it is. out required answer.
RAVI said:
1 decade ago
Can also be written as:
No.of ways (at least one boy) = Total no.of ways - No.of ways with no boy = 10c4 - 4c4 = 210 - 1 = 209.
No.of ways (at least one boy) = Total no.of ways - No.of ways with no boy = 10c4 - 4c4 = 210 - 1 = 209.
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