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How many arrangements of the letters in DIGITAL have two consecutive I’s?

I know this is a type combination, permutation problem but i'm a little unclear how to start with this problem.

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Hint: Glue the I's together to make a single "letter."

A less nice way: The position of the left-hand $I$ can be chosen in $6$ ways. Once this position is chosen, the position of the other I is determined. That leaves $5$ empty slots, which can be filled with distinct letters chosen from D, G, T, A, L in ??? ways.

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The answer is $6!$. The quick way, we have $6$ "letters" (counting the double I as a single letter), so $6!$ or $\binom{6}{1}\binom{5}{1}\cdots \binom{1}{1}$ (same). The longer way gives $6(5!)=6!$. Your calculation gives all $7$-letter words using our letters, and forgets about the consecutive I's condition. – André Nicolas Apr 19 '14 at 6:59

If we look at it like this: the two 'I's are always going to be together so we treat them as a single letter. So now we can calculate the possible arrangements of DIGTAL that is 6!.

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