Probability Question?

Question

Suppose a lottery were run by choosing five random numbered balls, from a pool of 90 balls numbered 1 - 90, how many possible groupings of fives balls can you have?

How od you solve this, on a calculator or by hand.

Answer 1

I'm going to assume that the order the balls are drawn in doesn't matter, since that's the way lotteries usually work. Here's the general formula for "combinations". It answers the question: "How many possible combinations are there of n things drawn k at a time (where k < n obviously). The answer is given by a "binomial coefficient, which is written as parentheses surrounding a column of two numbers - n on the top and k on the bottom. People usually refer to it as "n choose k" and it's equal to (n!)/[(n-k)! k!]. Where "k!" is k factorial and equals the product k(k-1)(k-2)...(3)(2).

The answer for "90 choose 5" (your problem) is: 43,949,268.

Answer 2

It depends on whether you put the balls back in after you pick them, so there's a chance that the same number can be picked more than once. Most lotteries don't put the balls back in, I think, so the probability would look like this:
90 possibilities for the first ball
89 possibilities for the second because you took one out
88 for the third, etc.
So it's 90 * 89 * 88 * 87 * 86 = 5273912160 different combinations of balls.

Answer 3

If you don't put the ball back into the pool, then it would be 90 x 89 x 88 x 87 x 86. If you put the ball back in then it would be 90 to the fifth power.

Answer 4

I t-h-i-n-k 90 X 5 = 450 (90 numbers, five balls)