Choose 11 Football Players in 12,367 WaysDate: 9/16/96 at 0:29:5 From: Anonymous Subject: Choose 11 Football Players in 12,367 Ways My question comes from the book "Discrete and Combinatorial Mathematics": p.30 No. 16. A gym coach must select 11 seniors to play on a football team. If he can make his selection in 12,376 ways, how many seniors are eligible to play? My process: nCr n! ------- = 12,376 r!(n-r)! n! ------- = 12,376 11!(n-11)! n! ------- = 4.940103168 e11 (n-11)! The problem is that we don't know of any way to negate the factorials; we can't distribute them, or anything else that we can think of. We know that the answer is n = 17, but we got that by trial and error on the calculator... so is there an actual method for this? Thank you in advance for your help. DC Date: 9/16/96 at 14:11:7 From: Doctor Tom Subject: Re: Choose 11 Football Players in 12,367 Ways Trial and error, in this case, isn't a bad method. Especially since you can zero in on it. You know the answer is more than 11, so try, perhaps, 20. If 20 is too small, try 30, and so on until you get an upper bound. But when you try 20, you'll find that gives too big an answer, so try something between 11 and 20, 15, say. That's too small so try 17 or 18, and so on. You can solve it directly, but it would be very ugly. Notice that n!/(n-11)! will completely cancel except for: n(n-1)(n-2)...(n-10) = 4.940...e11 This is an 11th degree polynomial if you multiply it out which you can solve for n. I think you'll agree that trial and error is faster. Actually, another thing you could do would be to factor 12376 to get 12376 = 2*2*2*7*13*17 Since there's a 17 in there, you know n is at least 17, and since there's no 19, it's less than 19, so 17 and 18 are the only two possibilities. -Doctor Tom, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/