Aptitude - Permutation and Combination - Discussion

At least 1 boy = total selections - no boy.

= 10c4-(6c0*4c4).
= 210-1.
= 209.

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?

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.

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.

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.

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.

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.

Is there anyone to explain why this particular question cannot be solved in the same method which were solved above?

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).

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?