Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer
    
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 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!.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.