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 6 of 10.
Atul said:
1 decade ago
I am not satisfied from the above ways because I don want to use simpler method.
Tell me why did we apply formula only on 1st and last terms. Why not on other terms?
Tell me why did we apply formula only on 1st and last terms. Why not on other terms?
Srinivas said:
1 decade ago
Can any one explain how we apply formula ncr = nc(n-r) to only first and last term only?
Mayur said:
1 decade ago
One Question has many times came no one gave answer.
Why formula use for only first term ncr(n-r)?
Why not for other one, how can we know when we use this formula?
Why formula use for only first term ncr(n-r)?
Why not for other one, how can we know when we use this formula?
Maggi said:
1 decade ago
In 1st step :........+ (6c1 x 4c3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4).
In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on.
In 2nd step :........+ (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
If we use formula nC(n-r) then it is not applied to 6C2 x 4C2. So on.
Shilpa said:
1 decade ago
6c2= 6!/2!(6-2)!
= 6!/2!*4!
= 6*5*4! 2*4!
Then 4! will be cancel.
= 6*5/2.
By solving it.
= 15 got in result.
= 6!/2!*4!
= 6*5*4! 2*4!
Then 4! will be cancel.
= 6*5/2.
By solving it.
= 15 got in result.
Shreu said:
1 decade ago
But the also there the children how it is possible?
Aditi Muley said:
1 decade ago
In third step last 1 is 6C2, I am confused in the fourth step about how we solved 6C2 that we got 15 in result.
Hitesh said:
1 decade ago
What about when there was asked for at most 2 boys?
Vijay 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).
In 1st step : ... + (6c1 x 4c3) +...6C4
In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?
Why don't the formula nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1 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).
In 1st step : ... + (6c1 x 4c3) +...6C4
In 2nd step : ... + (6c1 x 4c1) +...6C2 How ?
Why don't the formula nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1 apply to the other terms. Please Explain?
Prabhu 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?
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