# Probability that 5 different faces come up twice each if 6 side die is rolled ten times?

Find the probability that 5 different faces come up twice each if 6 side die is rolled ten times?

What methods should I apply here?

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Hint:

First, how many ways are there to pick 5 of the 6 faces.

If I have chosen 5 different faces, how many different orders can they come in. What is the probability that they come in that order.

Now multiply everything together.

From this I get: $$\frac{10!}{2^{5} 6^9}$$

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where did the 10!/2^10 come from? –  meiryo May 24 '11 at 0:46
@Meiryo: I was counting how many ways to arrange a,a,b,b,c,c,d,d,e,e in a row. Well there are 10! total, but we divide by 2! for each pair. So 10!/2^5. Aha! it seems my number needs to be modified! –  Eric Naslund May 24 '11 at 0:48
Ahh, I see now, thanks! –  meiryo May 24 '11 at 0:52

Just for kicks, here is a fun R simulation.

Running 10 repetitions of a 10,000 simulation run, we get an avg of 0.01178 ± 0.001

#  PARAMETERS:   #
# -------------- #
set.seed(1)
sides <- 6
rolls <- 10

# simulations to get a probability
simulations <- 10000

# repetitions to repeat the process, to get an average result
reps <- 10

#  ROLL THE DIE  #
# -------------- #
# This function is a single 10-roll sample
OneTrial <- function(sides, rolls) {
outcome <- sample(sides, rolls, TRUE)

# count the number of sides that came up exactly-two
#    and check if that amounts to 5 faces
sum(table(outcome) == 2) == 5
}

# --------------- #
#  MULTIPLE REPS  #
# --------------- #
results <- c()
for (i in 1:reps) {
simulOutcomes <- replicate(simulations, OneTrial(sides, rolls))
results[[i]]  <- sum(simulOutcomes) / simulations
}

mean(results)

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Less elegantly, the places for the $1$'s can be chosen in $\binom{10}{2}$ ways. For each such choice the places for the $2$'s can be chosen in $\binom{8}{2}$ ways, and so on. –  André Nicolas May 24 '11 at 1:33