- 03 Oct '09 00:01.... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:

"A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combination. I.E. How many nights will that one fairytale last?"

As said, I got the answer, but used Excel to do it - please would somebody help me with a more technical way of attacking this problem.

Many thanks.

Surtism - 03 Oct '09 01:15

I believe the formula you seek is*Originally posted by Surtism***.... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:**

"A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combina ...[text shortened]... mebody help me with a more technical way of attacking this problem.

Many thanks.

Surtism

C(n,r) = (n!)/(r!*(n-r)!)

C(n,r) the number of combinations of "n" distinct objects taken "r" at a time.

so your problem

C(14,4) = 14!/(4!*(14-4)!)

= (14*13*12*11*10!)/(4!*10!)

=(14*13*12*11)/(4*3*2*1)

=1001 days - 03 Oct '09 04:17

I don't quite follow your statement, "i.e. How many nights will that one fairytale last?"*Originally posted by Surtism***.... Its been forever since I looked at this sort of problem and I had to use Excel to calcualte the answer which really annoyed me as I seem to recall a simpler process. The problem was thus:**

"A sultan has 14 children. He chooses to read a fairy tale to 4 children each night. How many combinations of 4 children are there without repeating any combina ...[text shortened]... mebody help me with a more technical way of attacking this problem.

Many thanks.

Surtism

If the requirement is that each child can only hear the fairytale once, then there is simply 14/4 = 3.5 ~= 4 nights that the fairytale will last.

If you did every combination, every child would hear the fairytale multiple times. - 05 Oct '09 22:00

Thanks very much, so in simple terms:*Originally posted by joe shmo***I believe the formula you seek is**

C(n,r) = (n!)/(r!*(n-r)!)

C(n,r) the number of combinations of "n" distinct objects taken "r" at a time.

so your problem

C(14,4) = 14!/(4!*(14-4)!)

= (14*13*12*11*10!)/(4!*10!)

=(14*13*12*11)/(4*3*2*1)

=1001 days

4/14 x 3/13 x 2/12 x 1/11 = 1/1001

How would it work for the probability of getting 3 numbers on the UK lottery? - 06 Oct '09 03:20

ahhh...i think you just replace "n" with how many numbers are on the ticket, and "r" would be "3" for your particular situation....*Originally posted by Surtism***Thanks very much, so in simple terms:**

4/14 x 3/13 x 2/12 x 1/11 = 1/1001

How would it work for the probability of getting 3 numbers on the UK lottery?

I get

49!/(3!*(49-3)!)=18424

so 1/18424

I could be wrong.