Intermediate: Not talking about Level. Part-I
How many words can be made using all of the letters of the word "INTERMEDIATE" if the words neither begin with the letter "I" nor end with the letter "E"?
The answer is 12549600.
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1 solution
make it clear please.. i dnt get u
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when counting no. of combinations starting with 'I',we are fixing an 'I' in the first place and thereby counting the no. of possible combinations in the other places,similar is the case while counting combinations(of letters) ending with 'E'. After that inclusion-exclusion principle is applied i.e., |AUB|=|A| + |B| - |A intersection B|. Subtracting this result from total,we get our required result...hope i'm clear this time..:) if not,pls do ask..
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yep.. i gt..small mstakes.. well done!!thank you
Oh! i just forgot to subtract the cases of (and) from (or)..
I was trying to figure out for actual words. not just any random succession of letters. i fell like the question should be phrased a little differently. take out "words" and replace with "combinations of letters that do not start with I nor end with E" just my two cents. tough problem though!
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yes Sir,shld hav been comb of letters...i've found many such questions mentioning 'word' instead of group of letters,so i didn't get confused.
Why is my solution wrong? Here is my method:
Total cases when words don't start with with I or end with E =
1 0 ⋅ 1 0 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ⋅ 8 = 1 0 ! ⋅ 8 0
Removing repeated cases = 2 ! 2 ! 3 ! 1 0 ! ⋅ 8 0 = 1 2 0 9 6 0 0 0
Please help
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sry can't get u...i hv xplained elaborately..plz check,hope u'll make it out.
total possible combinations = 12!/(2!2!3!)=19958400... Starting with I = 11!/(2!3!)=3326400... Ending with E = 11!/(2!2!2!)=4989600.. Starting with I AND ending with E = 10!/(2!2!)= 907200.. hence, starting with I OR ending with E = (332600+4989600)-907200=7408800.. So, neither starting with I nor ending with E = 19958400-7408800=12549600 (Ans)