how can you distribute 8 SAME toys to 3 person? (it looks like an ordinary question but it is good..)
2006-10-13 03:44:56 · 7 answers · asked by alp 1 in Science & Mathematics ➔ Mathematics
(note that each children dont have to get at least one toy..)
2006-10-13 04:02:23 · update #1
still no correct results...
2006-10-13 04:12:06 · update #2
still no correct result
2006-10-13 04:25:24 · update #3
Hey, this is an interesting question.
First look at what happens with 2 children.
1 toy: 2 ways 01, 10
2 toys: 3 ways: 02, 11, 20
3 toys: 4 ways 03, 12, 21, 30
...
For 3 children you can give 8 to the first, and there is 1 way to divide the remainder among the other 2.
You can give 7 to the first and there are 2 ways to divide the 1 remaining gift between the other 2.
You can give 6 to the first and there are 3 ways to divide the 2 remaining
...
you can give 0 to the first and there are 9 ways to divide the 8 remaining
1 + 2 + ... + 9 = 45.
Hmmm, the sum of the first 9 integers... That looks like a Pascal's triangle (binomial coefficients). Without going through the details here, this problem can be generalized to splitting N gifts among M kids, and the answer will be (N+2, M) where () signifies the binomial coefficient.
2006-10-13 04:18:34 · answer #1 · answered by sofarsogood 5 · 1⤊ 0⤋
If you give 0 toys to the first person, you can divide the remaining 8 between the other two 8 different ways (0 and 8, 1 and 7, ..., 8 and 0). If you give 1 toy to the first person, you can divide the remaining 7 between the other two 7 different ways. And so forth until you can give 8 toys to the first person and have only 1 way to give 0 to each of the other two perople. So the number of ways is 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36.
Since the toys are the same, it doesn't matter which ones go to which people. I'm assuming that the people are distinguishable from one another. This isn't really a combination question. It's more about set theory or arithmetic.
2006-10-13 04:07:42 · answer #2 · answered by DavidK93 7 · 1⤊ 0⤋
right column represents # of permutations of the 3 numbers to the left:
8 0 0; 3
7 1 0; 6
6 2 0; 6
6 1 1; 3
5 3 0; 6
5 2 1; 6
4 4 0; 3
4 3 1; 6
4 2 2; 3
3 3 2; 3
total in rightmost column = 45; if correct admittedly not very elegant.
2006-10-13 04:57:32 · answer #3 · answered by Joe C 3 · 0⤊ 0⤋
choose 3 from 8 .....
8 C 3 = 8! / 5! * 3!
8 * 7 = 56 different ways of distributing all 8 toys
2006-10-13 04:00:49 · answer #4 · answered by Brian D 5 · 0⤊ 0⤋
This is the combination of 8 things, taken 3 at a time, or
8C3 Look up the formula and memorize it.
The answer should be 56
2006-10-13 07:06:47 · answer #5 · answered by Anonymous · 0⤊ 0⤋
give 1-1-6
simple
2006-10-13 03:54:49 · answer #6 · answered by saurabh s 1 · 0⤊ 0⤋
8C3 = 8!/(5!)
2006-10-13 05:50:09 · answer #7 · answered by Eyad E 3 · 0⤊ 0⤋