Evaluating binomial coefficiants

• Feb 25th 2014, 09:11 PM
crownvicman
Evaluating binomial coefficiants
Hi, I have three binomial coefficient problems that I need to prove.

a.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 31)

b.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 63)

c.) let a= (99 Choose 4) and let b = (100 choose 95) in terms of a and b.

I know a.) is less then b.) is equal to, and c.) is a+b simply by plugging them into a calculator but is there away to show these are correct without computations? My professor wasn't concerned with the actual values, more so he was looking for the reasoning behind why these are correct without actually computing them. Thank you
• Feb 25th 2014, 09:26 PM
romsek
Re: Evaluating binomial coefficiants
Quote:

Originally Posted by crownvicman
Hi, I have three binomial coefficient problems that I need to prove.

a.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 31)

b.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 63)

c.) let a= (99 Choose 4) and let b = (100 choose 95) in terms of a and b.

I know a.) is less then b.) is equal to, and c.) is a+b simply by plugging them into a calculator but is there away to show these are correct without computations? My professor wasn't concerned with the actual values, more so he was looking for the reasoning behind why these are correct without actually computing them. Thank you

Choose(n,k) = Choose(n,n-k)
Choose(n,k) is maximum at (n/2) if n is even or at (n-1)/2 and (n+1)/2 if odd
Choose(n,k) is monotonic increasing to this maximum and monotonic decreasing after it

so, for a given n, the farther away k is from the midpoint the smaller it is

That answers (a) and (b) w/o computation.

I don't quite get (c). Are you saying you want to express Choose(100,95) in terms of Choose(99,4) ?
• Feb 26th 2014, 02:28 PM
crownvicman
Re: Evaluating binomial coefficiants
Thank you very much. Sorry I mis-wrote question c.) it's supposed to be a= (99 choose 4) and let b= (99 choose 5). The question was to express (100 choose 95) in terms of a and b, which I'm still not quite sure how to explain that
• Feb 26th 2014, 02:44 PM
romsek
Re: Evaluating binomial coefficiants
Quote:

Originally Posted by crownvicman
Thank you very much. Sorry I mis-wrote question c.) it's supposed to be a= (99 choose 4) and let b= (99 choose 5). The question was to express (100 choose 95) in terms of a and b, which I'm still not quite sure how to explain that

$\left(\begin {array}{c}n \\ k \end{array}\right)=\left(\begin {array}{c}n-1 \\ k-1 \end{array}\right) + \left(\begin {array}{c}n-1 \\ k \end{array}\right)$
• Feb 26th 2014, 04:17 PM
Prove It
Re: Evaluating binomial coefficiants
Quote:

Originally Posted by crownvicman
Hi, I have three binomial coefficient problems that I need to prove.

a.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 31)

b.) Is (93 Choose 30) greater than, equal to, or less than (93 Choose 63)

c.) let a= (99 Choose 4) and let b = (100 choose 95) in terms of a and b.

I know a.) is less then b.) is equal to, and c.) is a+b simply by plugging them into a calculator but is there away to show these are correct without computations? My professor wasn't concerned with the actual values, more so he was looking for the reasoning behind why these are correct without actually computing them. Thank you

For a) it would help to remember that \displaystyle \begin{align*} {n\choose{k}} = \frac{n!}{k! \left( n - k \right) !} \end{align*}, which means \displaystyle \begin{align*} {93\choose{30}} = \frac{93!}{30! \cdot 63!} \end{align*} and \displaystyle \begin{align*} {93\choose{31}} = \frac{93!}{31! \cdot 62! } \end{align*}. Which of these is greater?

You can do the same for b).