I roll a fair 6-sided die five times. How many sequences of rolls will give me three 2's, one 4, and one 6? (A "sequence of rolls" is an ord

hi Chris,

There are 5 items so they can be arranged 5! ways.        $${\mathtt{5}}{!} = {\mathtt{120}}$$

But there are 3 that are the same.  I shall show you by example.

I have called the 3s       3A, 3B and 3C

Here are 3! = 6 of the  120 permutations

3A, 3B 3C 2 6

3A, 3C 3B 2 6

3B, 3A 3C 2 6

3B, 3C 3A 2 6

3C, 3A 3B 2 6

3C, 3B 3A 2 6

BUT each of these are really the same!   because the 3s are really identical.

so the number of permutations is       5!/3!

Is that all clear now?

I roll a fair 6-sided die five times. How many sequences of rolls will give me three 2's, one 4, and one 6? (A "sequence of rolls" is an ordered list of rolls, like 1, 1, 2, 1, 2.)

There are five positions where the 6 could appear.  For each of those there are four positions where the 4 could appear.  The 2's fill in the other positions.  So there are 5*4 = 20 sequences with one 6, one 4 and three 2's.

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what happened to my answer?   It has disappeared!  I must look under the bed!!!

I did have  5!/3! = 20

how annoying!

I wondered why you'd just repeated the question Melody!

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hi Alan,

I often repeat the question first but I don't know why my calculation wasn't there as well.

Just another of life's mysteries I guess.  :))

Melody, could you elaborate a little more on your answer??.....I'm having a "CDD" moment on this one....!!!!

Got it...it's like the number of identifiable five-letter words that can be formed when 3 of the letters are the same....

Thanks, Melody.....!!!

Yes it is exactly the same :)