43 presidents if you pick 5 how many groupings can you get
2007-04-26 16:55:56 · 4 answers · asked by Antonio 2 in Science & Mathematics ➔ Mathematics
i have the answer its 33,???
its 33 thousand i just cant figure out how they got it
2007-04-26 17:04:16 · update #1
remember this and remember this good.
selecting m objects from a set of N objests is calculated as a combination using the formula:
nCm. nCm = n!/m!*(n-m)!
43C5 = 43!/5!*38!
= 43 * 42 * 41 * 40 * 39/120 = 962,598 unique combinations.
2007-04-26 17:08:41 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Assume you can only pick a president once (the no replacement problem)
Combos = 43!/5!38! = 43 x 42 x 41 x 40 x39/ 2x3x4x5 = 43 x 7 x 41 x 2 x 39
2007-04-26 17:00:42 · answer #2 · answered by cattbarf 7 · 0⤊ 0⤋
well use the formula of probability here...
n!/ r!(n-r)!
= 43! / 5!(43 - 5)!
= 43! / 5!.38!
=43.42.41.40.39.38! / 5!.38!
= 43.42.41.40.39 / 5!
= 962598
2007-04-26 17:14:21 · answer #3 · answered by gt_thegame 2 · 0⤊ 0⤋
(43 choose 5) = 43! / 5!*(43 - 5)! = 962 598
2007-04-26 17:01:06 · answer #4 · answered by neo85888 1 · 0⤊ 0⤋