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HOW MANY DIFFERENT ARRANGEMENTS CAN BE MADE WITH THE NUMBERS
28535852

asked
User Parand
by
7.4k points

1 Answer

1 vote

Answer:

1,680

Explanation:

8 positions with basically 8 choices.

that is 8! arrangements.

but 2 is there 2 times.

8 is there 2 times.

5 is there 3 times.

only 3 is a single digit.

so, we need to eliminate every arrangement, where the 2s trade places, where the 5s trade places, and where the 8s trade places, because they are the same numbers.

that means we have to divide the 8! by 2!, then again by 2!, and again by 3!.

8! / (2! × 2! × 3!) = 8! / 24 = 1,680

answered
User Zsawyer
by
7.7k points

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