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ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together
im not sure what i did wrong but ans is 2/7
Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently:
Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply:
Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives:
Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore:
I hope this helps! That was a tricky one, stumped me for a bit there!