2006-12-17 14:15:21 · 9 answers · asked by dynoair821 1 in Science & Mathematics ➔ Mathematics
Number of ways = (President can be chosen in) 9 ways x (Vice president can be chosen in) (9-1=) 8 ways (after presiden is elected only 8 persons are left) x (Secretary can be chosen in) (9-2=) 7 ways = 9x8x7 ways.
Another solution:
If each person can participate again, even after being chosen, total number of ways = 9x9x9 ways.
2006-12-17 14:25:42 · answer #1 · answered by Sheen 4 · 0⤊ 0⤋
The answer is 9*8*7 = 504 ways
It is not 9C3 = 84 because for each group of three chosen, there are 3! = 6 different ways to choose among them for the three offices.
2006-12-17 14:28:22 · answer #2 · answered by Northstar 7 · 0⤊ 0⤋
504
2006-12-17 14:24:54 · answer #3 · answered by mmturtle 5 · 0⤊ 1⤋
9 * 8 * 7 ways = 504 ways
2006-12-17 14:18:33 · answer #4 · answered by Tom :: Athier than Thou 6 · 1⤊ 1⤋
9*8*7=504
2006-12-17 14:28:11 · answer #5 · answered by yupchagee 7 · 0⤊ 1⤋
6! ordinary strategies to rearrange the lads 6! ordinary strategies to rearrange the ladies the two can circulate first there are 36 ordinary strategies to decide on pres, 35 for vice chairman, etc 9! (can no longer avert giving the respond) intercourse is beside the point 35 pres, 34 vice chairman, etc. after the officers, there are 30C5 ordinary strategies to %. the committee there are 2^5 plausible outcomes how lots of them have not any heads?
2016-10-15 03:48:10 · answer #6 · answered by ? 4 · 0⤊ 0⤋
This is a permutation problem.
P(n,r) = n(n-1)(n-2)...(n-r+1)
P(9,3)= 9 * 8 * 7
There are 9 choices for P, which leaves 8 choices for VP & 7 choices for S.
2006-12-17 14:37:10 · answer #7 · answered by S. B. 6 · 0⤊ 0⤋
I'd be inclined to say 9x8x7 = 504. Those are unique offices and we assume one person can't fill 2 offices.
2006-12-17 14:27:45 · answer #8 · answered by answerING 6 · 0⤊ 0⤋
no of ways=9C3=9*8*7/1*2*3
=84 ways
2006-12-17 14:17:43 · answer #9 · answered by raj 7 · 0⤊ 2⤋