How many ways are there to arrange "RACECAR" such that the word you make starts with "ACE"
Counting "ACE" as one part, we get there are $$\frac{5!}{2!} = 5*4*3 = 60$$
ways to arrange "RACECAR" so that "ACE" is together.
Is this correct?
How many ways to arrange the word?
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$\begingroup$
combinatorics
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0The question says "starts with ACE", not "contains ACE". – 2017-01-24
3 Answers
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since letters 'ACE' are fixed in beginning so the remaining letters A, C, R, R can be arranged by $4!/2!=12$
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Put $ACE$ at the beginning, then count the number of ways to arrange $\color\red{A}\color\green{C}\color\orange{RR}$:
$$\frac{(\color\red1+\color\green1+\color\orange2)!}{\color\red1!\times\color\green1!\times\color\orange2!}$$
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Words start with ACE. So ACE is fix. It should be done in only 1 way.
Remaining alphabets are RACR.
Now we have total 4 alphabets so 4 alphabets can be arrange in 4! ways.
Total ways = 1 × 4!
But R is repeating 2 times. So divide total ways by 2!
Total ways = $\frac{1 × 4!}{2!}$
= $\frac{1 × 4 × 3 × 2!}{2!}$
= 1 × 4 × 3 = 12