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In a casino, a person picks 4 numbers from a set of 9 numbers (3 Even, 6 Odd). A person wins if at least one of those 4 numbers is Even. What are the total ways he can Win ?

There are two approches I believe.

1) One is to find all possible ways of picking 4 from 9 (9C4) and subtract the number of ways of getting only odd (6C4) ie:

9C4 - 6C4

2) Pick one even from the set of 3 Evens and pick 3 from the remaning 8. ie :

3C1 * 8C3

What is the difference between the above two, In my oppinon the 2nd method does the same, why is it wronge ?

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    In your second method some cases are being counted more than twice. – 2017-01-01

2 Answers 2

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The second method is wrong because you are over counting.

Suppose the even numbers are $e_1,e_2,e_3$ and the odd numbers are $o_1,o_2,\cdots,o_6$.

When you choose a number from the set of three evens, suppose you get $e_1$. Out of the remaining $8$, you choose $3$, let them be $e_2,o_1,o_2$.

So you have $e_1,e_2,o_1,o_2$ and you win.

But you could also have selected them this way. When you chose one out the three evens, you got $e_2$. Later, when you chose the next three, you got $e_1,o_1,o_2$. The set of numbers at the end is identical.

However, your method counts these two cases separately, whereas they should be counted exactly once. So, the final answer you get is incorrect.

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The problem is that in the second method you are marking one specific odd element in each set. So the combinations which have more than one odd number are counted more than once.

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    Actually it is not missing combinations but counting some combinations more than twice. – 2017-01-01
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    oh, I thought he was picking the other elements from the odd ones. – 2017-01-01
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    sorry for the confusion, thanks for the help @WiCK3DPOiSON – 2017-01-01