A magic show has got ten people within the audience. In order to perform the next act, the magician requires four people to help from the audience. In how many different ways can the magician ask four people from the audience (of ten people) to come up to the stage for help?
Question
A magic show has got ten people within the audience. In order to perform the next act, the magician requires four people to help from the audience. In how many different ways can the magician ask four people from the audience (of ten people) to come up to the stage for help?
Solution
This is a combination problem. The magician doesn't care about the order in which the people come up, just who they are. So, we use the combination formula:
C(n, r) = n! / [r!(n-r)!]
where:
- n is the total number of options,
- r is the number of options chosen at a time,
- "!" denotes a factorial, which means multiplying all positive integers up to that number.
In this case, n = 10 (the total number of people in the audience) and r = 4 (the number of people the magician needs).
So, the calculation would be:
C(10, 4) = 10! / [4!(10-4)!] = 10! / [4! * 6!] = (10 * 9 * 8 * 7 * 6!) / (4 * 3 * 2 * 1 * 6!) = (10 * 9 * 8 * 7) / (4 * 3 * 2 * 1) = 210
So, the magician can choose 4 people out of 10 in 210 different ways.
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