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Combinatorics, please help!

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by maihuna » Sun Jan 23, 2011 3:09 am
5 vechiles out of 10 can be choosen in 10c5 way

Out of such no bikes in 6c5 ways

so min 1&1 combo in 10c5-6c5
= 10*9*8*7*6/5*4*3*2 - 6
= 42-6
= 36
Charged up again to beat the beast :)
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by uriall » Sun Jan 23, 2011 3:19 am
maihuna wrote:5 vechiles out of 10 can be choosen in 10c5 way

Out of such no bikes in 6c5 ways

so min 1&1 combo in 10c5-6c5
= 10*9*8*7*6/5*4*3*2 - 6
= 42-6
= 36
Thank you! it really helped!
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by Rahul@gurome » Sun Jan 23, 2011 3:22 am
uriall wrote:If you have 4 bikes and 6 cars, how many combinations of 5 vehicles can you make if every combination of the 5 vehicles must at least contain 1 bike and 1 car?
Out of 10 vehicles, there are 6 cars.
Hence if you select any 5 cars out of these 10, you'll always have at least 1 car in your selection.

Now number of possible selections of 5 vehicles out of 10 = 10C5 = 252
All of these 252 selections has at least 1 car in them.

Now in how many selections we don't have no bike at all?
This is only possible if all of the 5 vehicles are selected out of 6 cars.
Which can be done in 6C5 = 6 ways.

Hence, number of selections of 5 vehicles out of 10 such that there is at least 1 bike and 1 car = (252 - 6) = 246
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by Rahul@gurome » Sun Jan 23, 2011 3:25 am
uriall wrote:....
10c5-6c5
= 10*9*8*7*6/5*4*3*2 - 6
= 42-6
= 36
Check the calculation again.
You're missing a 6 there.
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by uriall » Sun Jan 23, 2011 3:28 am
Rahul@gurome wrote:
uriall wrote:....
10c5-6c5
= 10*9*8*7*6/5*4*3*2 - 6
= 42-6
= 36
Check the calculation again.
You're missing a 6 there.
You are right!
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