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### Choose 11 Football Players in 12,367 Ways

```
Date: 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/
```
Associated Topics:
High School Permutations and Combinations

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