Derek wants to display 6 of the marbles in his collection in 2 different showcases. If each showcase must have at least 1 marble, how many different ways can he display the marbles?

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Mathematics

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10 September, 16:13

Derek wants to display 6 of the marbles in his collection in 2 different showcases. If each showcase must have at least 1 marble, how many different ways can he display the marbles?

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Answers (2)

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Chris Weiss · Answer 1

6

Step-by-step explanation:

This is a case of combination since the order does not matter, it only matters that there must be a marble in each one inside, therefore we must apply its formula.

xCn = n! / x! * (n - x) !

n, in this case it would be 4 because we must subtract 2 from the total 6, since there must already be one inside in each display case.

x = 2, since it is the number of showcases.

Replacing:

2C4 = 4! / 2! * (4 - 2) ! = 6

Which means that you can in six different ways.

Scarlett Mccann · Answer 2

6

Step-by-step explanation:

This is going to be solved using the combination formula [nCr = n! / r! * (n - r) !, where n represents number of total items, r represents number of items chosen at that time], we know this because the order is irrelevant but selection is important. There are already one marble in each showcase, so we only have 4 marbles to work with.

It then becomes

4C2 = 4! / (4-2) ! * 2!

= 6