Exactly 4 actors try out for the 4 parts in a play. if each actor can perform any one part and no one will perform more than one part, how many different assignments of actors are possible?
the answer is 24 but I don't know how to solve it. And please explain it to me like I'm an elementary student.
thanks in advance
2006-11-18 06:52:31 · 3 answers · asked by Anonymous in Education & Reference ➔ Homework Help
There's two ways you can do it. The simple way is to multiply 1x2x3x4=24. Don't know why it always works, but it does.
The extended version? Say the actors are 1, 2, 3, and 4, and the parts are A, B, C, and D. These are the possible combinations:
1A, 2B, 3C, 4D
1A, 2B, 3D, 4C
1A, 2C, 3B, 4D
1A, 2C, 3D, 4B
1A, 2D, 3C, 4B
1A, 2D, 3B, 4D
...those are the combinations for if 1 was A. The same # of combinations are available for if 2, 3, and 4 were available, so 6x4=24.
2006-11-18 07:36:36 · answer #1 · answered by Anonymous · 0⤊ 0⤋
there are 24 variations of the four actors
1 - stays the same 234,243,324,342,423,432
6 variations for each* times 4 actors
2006-11-18 06:59:12 · answer #2 · answered by tomkat1528 5 · 0⤊ 0⤋
4x3=12 12x2=24 1x24=24
4x3x2x1 same if there are 8 possibilities...8x7x6x5x4x3x2 ( the 1 isn't really necessary.
2006-11-18 06:59:58 · answer #3 · answered by max2959 3 · 0⤊ 0⤋