combinations & permutations

[carebear]

I am almost done this unit. I hope this is my last question. Thank you very much for all of the help that you all have given me so far. It is greatly appreciated.

How many four letter arrangements can be made from the letters DDEFGH?

This is my answer:
Case 1 ("repeats"): C(2,2) x C(4,2) x 4!/2! = 72
Case 2: ("no repeats"): P(5,4) = 120
72 + 120 = 192 arrangements

My friend did it this way:
Case 1: (both D's): 4!/2! = 12
Case 2: (1 D): 4! x 2 = 48
Case 3: (no D's) 4! = 24
12+48+24 = 84 arrangements

Who is right? I feel both ways should work....why don't they? Or did one of us make an error with our cases?

Thanks again!

 

[galactus]

Your friend undercounted. You are correct.

See your friends mistake(s)?.

i.e. For their first case, they did not count the number of ways of choosing the 4 letters from 6. They just counted the number of arrangements of 4 letters with two repeats.

Case 1, 2 D's: place the two D's and choose 2 from the remaining non-D's in 6 ways. Then, arrange in 4!/2=12 ways. 12*6=72

Case 2, 1 D: place a D and choose 3 from the 4 non-D's in 4 ways. 4*24=96

Case 3: no D's: arrnage the 4 non-D's in 24 ways.

72+96+24=192

 

[Denis]

My friend did it this way:

You got a friend ? :shock: