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Source: — Problem Solving |
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by sudhir3127 » Tue Aug 12, 2008 8:30 pm
12C2 = 66
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by medea66 » Wed Aug 13, 2008 6:14 pm
Forgot about combinations, thanks!
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Detailed

by goldenpikapp » Sun Mar 08, 2009 5:35 pm
12 x 11 x 10 x 9 x 8 x ect divided by 2! x 10 x 9 x 8 x ect.

Cancel out the 10 x 9 x 8 x ect. and you are left with

12 x 11 divided by 2! or 132 divided by 2 = 66
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by vittalgmat » Sun Mar 08, 2009 9:24 pm
medea66 wrote:Forgot about combinations, thanks!
Hi Medea66,
Combinations is the simplest way to handle this problem. Here are my thoughts on how I would approach this problem.

1. The problem asked how many ways.. blah
So my first question is it a permutation or a combination ?
Here this is not an arrangement. Picking up bulb #3 and then bulb #4 is same as picking up bulb #4 and then picking up bulb #3.
Hence Combinations it the way to go.

2. I dint see any restrictions imposed. Hence is a simple combination problem.

The problem boils down to "in how many ways can I select 2 bulbs from a a box of 12".

soln is 12C2 = 66

This is a vanilla combinations problem in the low 500 range IMO. So u can expect this type of problem where u directly apply the formula.

HT Helps.
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