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Total number of Arrangement of the words SUCCESS in which the consonants appear in

Alphabetic order, is

My Try:

Total no. of Arrangement of the word SUCCESS Taken all at a time $\displaystyle = \frac{7!}{2!\times 3!} = 420$

Now Arrangement of the word which Contain only CCSSS taken all at a time $\displaystyle =\frac{5!}{2!\times 3!} = 10$

Now I did not Understand How can I solve it.

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up vote 11 down vote accepted

As you've shown, there are 420 ways to arrange the word SUCCESS, and there are 10 ways to arrange the consonants CCSSS.

Of these 10 ways, only one has the consonants in alphabetical order (namely, CCSSS); so by symmetry the number of arrangements of SUCCESS with the consonants in alphabetical order will be $\frac{1}{10}(420)=42$.

An alternate way to do this, similar to the other answers, is to first arrange the consonants in alphabetical order: CCSSS.

Then we have $\binom{7}{2}$ ways to choose blank spaces where E and U will go, since there are 2 letters and 5 dividers (C,C,S,S,S), and 2 ways to choose in which of the two places the E will go, so there are $\binom{7}{2}\cdot2=42$ arrangements.

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One say to think of it is to write the letters C and S like this: C C S S S

The letter E can be inserted into any of the $6$ "gaps" (these include the $2$ endgaps). And for every way to insert the E, there are $7$ places to insert the U.

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You can replace all the consonants with 'X" without affecting the answer, since their order is prescribed.

So how many arrangements of 'XUXXEXX' are there? Start with 'XXXXX'. There are 6 possitions you can stick in a 'U', then there are 7 positions you can stick in a 'E'. That makes 42 total arrangments.

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or $\frac{7!}{5!}=42$. – user26486 Sep 4 '14 at 21:13

We know that the consonants have to be CCSSS. So let the U go in front: UCCSSS. Where can the E go? It could go before the U, EUCCSSS; immediately after, UECCSSS; after the first C, UCECSSS; and so on until UCCSSSE. There are 7 such arrangements. Now no matter where we put the U there will be 7 arrangements (try a few and you'll see it's obviously so) and there are 6 places to put the U. $6 \times 7=42$ arrangements.

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