5 red balls, 2 white balls and 3 blue balls are arranged in a row. If the balls of the same color are not distinguished from each other. How many possible ways can they be ordered?

So let's assume that in any arrangement we are not able to distinguish the balls having the same colour, so only different orders of the colours involved are counted.

If we number these balls we have 12 different numbers, 3 of which belong to the white balls, 4 belong to the blue balls and the final 5 belong to the red balls. They could be arranged in 12! different ways. Pick any of these arrangements.

The white balls in this arrangement are now ordered because I gave these balls a number. But I specifically stated in the first alinea that we were not interested in all these possible orders of the white balls within a sequence of 12 balls. So we must correct for these orders, there are 3! of these, each one leading to exactly the same visual order of white balls. The other colours lead to a correction of 4! and 5!. Moreover, these corrections are independent of each other. I can change the position of the white balls without interrupting the order of the colours of the remaining balls. So we should divide by the product of 3!,4! and 5!.

PiaDeveau PiaDeveau · Answer 2 · 2022-03-17T17:38:01+01:00

Total number of arrangement = 3,628,800

Arrangement based problem:

Given that;

Number of red ball = 5

Number of white ball = 2

Number of blur ball = 3

Find:

Total number of arrangement

Computation:

Total number of balls = 5 + 2 + 3

Total number of balls = 10

Total number of arrangement = 10!

Total number of arrangement = 3,628,800

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