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I just started a course about Statistics and I was wondering if I am doing the following question right.

Q: A student enters 2 of 6 positive numbers with "-" signs and 3 of 4 negative numbers with "+" signs. If at some stage, a program chooses 3 distinct numbers from these 10 random numbers without replacement and multiplies them, we want to calculate the probability that at this stage no error occurs.

Totally, how many different ways are there of choosing three numbers for the program at that stage?

My answer:

So, just say I entered these 10 numbers

-1 -2 3 4 5 6 7 8 9 -1, but I really intended for 1 2 3 4 5 6 -7 -8 -9 -1. Therefore, there are in total 5 numbers that are entered incorrectly. In order for no error, the computer must choose 3 numbers from a possible 5 numbers that were entered correctly. Therefore, the number of ways would be

5 choose 3 = 5! / (3! (5 - 3)!) = 10 ways correct?

Thanks for your time.

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# Combinations (n choose r)

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