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?

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- Nov 23rd 2008, 02:18 PMcnmath16Combinations
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 PMAryth
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:

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

For r=0:

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

$\displaystyle = 1$

For r=1:

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

$\displaystyle = 5$

For r=2:

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

$\displaystyle = 10$

For r=3:

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

$\displaystyle = 10$

For r=4:

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

$\displaystyle = 5$

For r=5:

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

$\displaystyle = 1$

Now you add them all together:

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

That tells you that there are 32 possible combinations