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).
![]() of 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 4 of 10.
Alex said:
1 decade ago
I completely don't get it! Combintations:
1B+3G
2B+2G
3B+1G
4B
Those are very clear, no doubts.
so, the final answer is (1b+3g)+(3b+2g)+(3b+2g)+4b
Let's see the first one.
1b+3g=BGGG (as we have 4 spots). So, for the boy we have 6 different options (as any of the 6 boys can be chosen). We have 3 spots for girls and 4 girls to fill those 3 spots. So, for the first of the 3 girl spots there are 4 different options to choose from (as we have 4 girls), for the second spot we have only 3 girls left (as the forth girl already took the first spot), and for the 3d spot we have 2 girls left. So, the total result for 1b+3g would be 6x4x3x2+ we have to account the other combinations (2b+2g, 3b+1g, 4b). So, how the result is only 209?
1B+3G
2B+2G
3B+1G
4B
Those are very clear, no doubts.
so, the final answer is (1b+3g)+(3b+2g)+(3b+2g)+4b
Let's see the first one.
1b+3g=BGGG (as we have 4 spots). So, for the boy we have 6 different options (as any of the 6 boys can be chosen). We have 3 spots for girls and 4 girls to fill those 3 spots. So, for the first of the 3 girl spots there are 4 different options to choose from (as we have 4 girls), for the second spot we have only 3 girls left (as the forth girl already took the first spot), and for the 3d spot we have 2 girls left. So, the total result for 1b+3g would be 6x4x3x2+ we have to account the other combinations (2b+2g, 3b+1g, 4b). So, how the result is only 209?
Nikhil said:
1 decade ago
At least 1 boy = total selections - no boy.
= 10c4-(6c0*4c4).
= 210-1.
= 209.
= 10c4-(6c0*4c4).
= 210-1.
= 209.
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?
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?
Hitesh said:
1 decade ago
What about when there was asked for at most 2 boys?
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.
Shreu said:
1 decade ago
But the also there the children how it is possible?
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.
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.
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?
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