In a certain television program there are 40 bags, 15 to be chosen if the participant answers correctly (but suppose he gets them all right) out of those 40 there are 8 bags that will remove the high prizes. So what is the chance that I will choose 15 correct bags?
I thought that way:
P = probability of the correct bag
P(first bag) = 32/40, P(second bag) = 31/39 and so on...
In this logic the answer would be:
(32!/(32-15)!)/(40!/(40-15)!) = 0.01406383759 or 1.4%
But I have a feeling that this is wrong because I learned at school that when the order doesn't matter we use the combination formula, I hope someone will answer me if I'm correct :)
$\endgroup$2 Answers
$\begingroup$You are correct in a sense. We have to formulas to consider and compare:$$ nPr=\frac{n!}{(n-r)!}\text{ and }nCr=\frac{n!}{r!(n-r)!} $$but when you divide two of those using different values of $n$ but the same value of $r$, the two results will be equal whether using combinations (nCr) or permutations (nPr).
I will give you two reasons for this:
- Suppose we give the good bags an ID 1 through 32 and the bad ones 33 through 40. Then you could consider a successful outcome either as an ordered sequence from the 32 good bags, or just some subset of the 32 bags without considering their order. For the ordered case, to compute the probability of success, you would then have to divide by ALL ordered sequences from the 40 bags. Hence $32P15/40P15$. But if you only considered groups of $15$ without considering order, it should be $32C15/40C15$.
- The factor $r!$ cancels out when dividing the combinations formulas with the same $r$.
They are two different ways to calculate the same probability.
$$\frac{\binom{32}{15}}{\binom{40}{15}} = \frac{\frac{32\times\dots\times18}{15!}}{\frac{40\times\dots\times26}{15!}} = \frac{32\times\dots\times18}{40\times\dots\times26} = \frac{32}{40} \times \frac{31}{39} \times \dots \times \frac{18}{26}$$
- LHS : number of correct combinations divided by number of possible combinations
- RHS : your thinking
Note that the $15!$ on both the denominator and the numerator cancel each other.
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