Question

An urn contains five balls labelled A, B, C, D, E. You draw three balls at random, without replacement. How many different combinations of three letters can you draw? Combination means that order of draw is not important. Example: ACE and CAE are counted as the same combination, but ABD, ABE, CDE, etc are different combinations.

Answer 1

Using the combination formula, it is found that you can draw 10 different combinations of three letters.

The order in which the letters are drawn is not important, as stated in the problem, ACE and CAE are counted as the same combination, hence the combination formula is used instead of the permutation formula.

What is the combination formula?

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula, involving factorials.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

For this problem, three balls are taken from a urn with five balls(labeled A, B, C, D and E), hence n = 5, x = 3 and:

$C_{5,3} = \frac{5!}{3!2!} = 10$

You can draw 10 different combinations of three letters.

More can be learned about the combination formula https://brainly.com/question/25821700

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