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  1. In how many ways can the letters of the word CHROMATIC be arranged,

a. without any restrictions

b. if the position of the vowels can’t change

c. find the probability that the string of letters begins with the letter M

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closed as off-topic by Graham Kemp, C. Falcon, Adam Hughes, pjs36, projectilemotion Mar 8 '17 at 15:47

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  • $\begingroup$ Well, tell us what you have tried and we might help. $\endgroup$ – Graham Kemp Mar 8 '17 at 2:58
  • $\begingroup$ I understand how to do this if each letter is distinct, but not so much if the letters are repeated $\endgroup$ – N. D Mar 8 '17 at 2:59
  • $\begingroup$ If letters are repeated, start by pretending they are all distinct. Then note that two orders with just the $C$s interchanged are identical, so you need to divide by $2$ because you have double counted. $\endgroup$ – Ross Millikan Mar 8 '17 at 3:14
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a. without any restrictions

I understand how to do this if each letter is distinct, but not so much if the letters are repeated

There are $9!$ distinct ways to rearrange $9$ unique letters, but here we have $2$ that are indistinguishable (the CC).   So every arrangement is one from $2!$ that are indistinguishable.   Thus the count we seek is: the ways to rearrange $9$ symbols when $2$ are indistinguishable. $$\dfrac{9!}{2!}$$

Alternatively: We can count ways to select $2$ from $9$ places to put the C, then arrange the remaining $7$ letters.   Same thing. $$\binom 92 7!$$

b. if the position of the vowels can’t change

There are two ways to read this.

1) That the three vowels cannot be moved.   If so, just count the rearrangements $6$ consonants when exactly $2$ are indistinguishable.

2) That the positions for the vowels are fixed but not their placement among them.   If so, multiply the above by the count for rearrangements of $3$ distinct vowels.

c. find the probability that the string of letters begins with the letter M.

The favoured outcomes have $M$ in first place and the remaining $8$ letters (of which exactly $2$ are indistinguishable) arranged in the remaining $8$ places.

You have already found the size of the sample space.   All arrangements are equally probable (are they not?), so the probability is the ration of favoured over total.

(Though there is a much easier way....think about it.)

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