Number of ways to arrange letters of word - confirm solution

[MaryStew]

How may distinct ways are there to rearrange the letters in PUNKEYDOODLES?

So at first I thought it'd be 13!/(2! x 2! x 2!)

but then I was following a solution for a similar question online and got 13C2 x 11C2 x 9C2 x 8C1 x 7C1 x 6C1 x 5C1 x 4C1 x 3C1 x 2C1 x 1C1 which gives a completely different answer.

Which solution would it follow? I'm very confused.

 

[pka]

You were correct the first time: $\displaystyle \frac{13!}{(2!)^3}=778377600$ see here

 

[Steven G]

The first one is correct.
You can arrange 13 letters in 13! ways. However if say one letter is repeated three times then you can write the same ONE ordering 3! ways. For example suppose the letter Q appears three time. Let's denote the three Qs as Q1, Q2 and Q3. Suppose the Qs were in positions 1, 3 and 7. Lets look at JUST those three positions. The 13 letter arrangement can have Q1 in the 1st position, Q2 in the 3rd position and Q3 in the 7th position OR it could have Q2 in the 1st position, Q1 in the 3rd position and Q3 in the 7 th position OR....
We can rearrange those Qs in 3! ways. Regardless of where the 3 Qs end up we can rearrange the Qs in 3! ways and still have the SAME ordering of the 13 letters (since there are NOT really Q1, Q2 and Q3 but rather Q, Q and Q). So we must divide the 13! by 3!. Is this clear?

 

[pka]

How may distinct ways are there to rearrange the letters in PUNKEYDOODLES?
but then I was following a solution for a similar question online and got 13C2 x 11C2 x 9C2 x 8C1 x 7C1 x 6C1 x 5C1 x 4C1 x 3C1 x 2C1 x 1C1 which gives a completely different answer.

See Here $\dbinom{13}{2}\dbinom{11}{2}\dbinom{9}{2}(7!)=778377600$ the two approaches give the same result.

 

[Romsek]

See Here $\dbinom{13}{2}\dbinom{11}{2}\dbinom{9}{2}(7!)=778377600$ the two approaches give the same result.

Since I think this is an important concept to grasp I'm going to expand on this equation.
We've got two O's, two D's, two E's, and the rest 7 single characters.

What pka's solution does is first slot the two O's. There are $\displaystyle \dbinom{13}{2}$ ways to do that.
We now have 11 slots left and we slot the two D's. There are $\displaystyle \dbinom{11}{2}$ ways to do that.
We now have 9 slots left and we slot the two E's. There are $\displaystyle \dbinom{9}{2}$ ways to do that.
We now have 7 slots left and 7 single characters. There are 7! ways to arrange them.

Combine all this and you get the result quoted above.

 

[JeffM]

How may distinct ways are there to rearrange the letters in PUNKEYDOODLES?

So at first I thought it'd be 13!/(2! x 2! x 2!)

but then I was following a solution for a similar question online and got 13C2 x 11C2 x 9C2 x 8C1 x 7C1 x 6C1 x 5C1 x 4C1 x 3C1 x 2C1 x 1C1 which gives a completely different answer.

Which solution would it follow? I'm very confused.

[MATH]\dbinom{13}{2} * \dbinom{11}{2} * \dbinom{9}{2} * \dbinom{7}{1} * \dbinom{6}{1} * \dbinom{5}{1} * \dbinom{4}{1} * \dbinom{3}{1} * \dbinom{2}{1} * \dbinom{1}{1} =[/MATH]
[MATH]\dfrac{13!}{2! * 11!} * \dfrac{11!}{2! * 9!} * \dfrac{9!}{2! * 7!} * 7 * 6 * 5 * 4 * 3 * 2 * 1 =[/MATH]
[MATH]\dfrac{13 * 12}{2} * \dfrac{11 * 10}{2} * \dfrac{9 * 8}{2} * 7! =[/MATH]
[MATH]\dfrac{13 * 12 * 11 * 10 * 9 * 8 * 7!}{2! * 2! * 2!} = \dfrac{13!}{2! * 2! * 2!}.[/MATH]
Your original method was valid, notationally more compact, and, with a good enough calculator, quicker to evaluate than the second method. That does not make the two methods logically inconsistent. They result in the same numerical value. If by one method you got an answer of
2 * 6 * 2 and by another method an answer of 8 * 3, would you say that you got "completely different" answers?

Frequently, there are many ways to skin a cat. This is true in combinatorics,, but there answers that look very different notationally may be identical numerically.