Can anyone help me with this question?
"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?"
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Combinatorics, please help!
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maihuna
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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
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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Rahul@gurome
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Out of 10 vehicles, there are 6 cars.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?
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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Rahul@gurome
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Check the calculation again.uriall wrote:....
10c5-6c5
= 10*9*8*7*6/5*4*3*2 - 6
= 42-6
= 36
You're missing a 6 there.
Rahul Lakhani
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