# Thread: Arrangements of SUCCESS

1. ## Arrangements of SUCCESS

To find how many arrangements contain the two C's next to each other, I considered them as a unit and got 6!/3!=720/6=120.
I actually wrote them out and counted them to verify.
To find out how many arrangements containing the two C's also contain two or more S's next to each other, I considered each pair as a unit and got 5!=120.

Obviously, this isn't correct. in my actual count I only found 96.

What is the correct way to find the answer?

Thanks.

2. ## Re: Arrangements of SUCCESS

SUCCESS = 1233411 ; easier...

Lowest arrangement: 1112334
highest: 4332111

3. ## Re: Arrangements of SUCCESS

Originally Posted by thayer
To find out how many arrangements containing the two C's also contain two or more S's next to each other, I considered each pair as a unit and got 5!=120.
I will assume that all seven letters and the C's need not be together.
Find the number of arrangements in which none of the S's are together:
$__U__C__C__E__$ there are four places to place the S's so as to separate them.
There are $\dbinom {4}{3}=4$ ways, the separators can be arranged in $\dfrac{4!}{2!}$ ways.

So the answer is $\dfrac{7!}{2!\cdot3!}-4\cdot\dfrac{4!}{2!}$ YOU MUST EXPLAIN WHY TO US.
Note: my answer does not assume the C's are together.
You wrote "containing the two C's also contain two or more S's next to each other,