Combinations

• Nov 23rd 2008, 02:18 PM
cnmath16
Combinations
A pharmacist uses 5 separate weights: 1 g, 2g, 4 g, 8 g and 16 g. If the pharmacist can combine these weights to create new weights, how many different weights are possible?
• Nov 23rd 2008, 06:44 PM
Aryth
Well, it's a simple problem. You could just find the possible combinations of picking differing sets of weight. Combinations of 1 weight picked from 5, combinations of 2 weights picked from 5, combinations of 3 weights picked from 5... Etc. You use the following formula for n being the total number of weights and r being the number of weights you're choosing:

$C^n_r = \frac{n!}{r!(n - r)!}$

For r=0:

$C^5_0 = \frac{5!}{0!(5)!}$

$= 1$

For r=1:

$C^5_1 = \frac{5!}{1!(4)!}$

$= 5$

For r=2:

$C^5_2 = \frac{5!}{2!(3)!}$

$= 10$

For r=3:

$C^5_3 = \frac{5!}{3!(2)!}$

$= 10$

For r=4:

$C^5_4 = \frac{5!}{4!(1)!}$

$= 5$

For r=5:

$C^5_5 = \frac{5!}{5!(0)!}$

$= 1$

Now you add them all together:

$Total = 1 + 5 + 10 + 10 + 5 + 1 = 32$

That tells you that there are 32 possible combinations