A quick question on combination and permatation

karkar athlete

Hi I have a question on a problem I have. I am always unsure whether it is combination or permatation.

But this is the question I am stuck on "How many arrangements of the letters in \MATHEMATICS" are there in which TH appear together but the TH is not immediately followed by an E (not THE)?".

I did it by hand and got 81 ways, but I am wondering is there some sort of formula either combination or permatation that I should be using as we are not meant to do it by hand, and I cant quite get 81 using combination and permatation methods. Also I am not 100% sure if that is even the write answer.

Thanks a lot for your help

hivizman

You are dealing with permutations here, because you are trying to count different arrangements of the letters subject to two constraints (the arrangement must contain the letters TH in that order but cannot contain the letters THE in that order).

If you didn't have any constraint, then you have to calculate all the arrangements of 11 letters, which is 11! (11 factorial). But in the letters there are two "A"s, two "M"s and two "T"s, and swapping the two "A"s (say) around in any arrangement simply gives the same arrangement. So you have to divide 11! by 2 x 2 x 2 = 8 to give the total number of arrangements of MATHEMATICS without any constraints.

How to deal with the constraints? I suggest that you think of "TH" as a single letter. This then means that the total number of arrangements drops to 10!/8 (again you have to adjust for the two "A"s and the two "M"s, and also for the two ways in which you can get "TH", given the two "T"s).

Then on the same argument the total number of arrangements containing the sequence "THE" is 9!/8.

So the number of arrangements that satisfy both constraints is:

10!/8 - 9!/8 = 9! x 9/8 (as 10! = 9! x 10)

This is 408,240.

Interestingly, this is your number of 81 multiplied by 5040 = 7! = 8!/8, so what I think you have done is treat this as a combination situation rather than a permutation. You need to treat each arrangement of the other 8 letters (other than "THE") separately (that gives the 8! term), but also allow for the duplicated letters (that gives the division by 8 term).

There are 10 positions for the "TH" in an 11-letter arrangement, and then 9 places for the "E", giving a total of 10 x 9 = 90 possibilities. But in 9 of these (i.e. all positions for "TH" except right at the end of the arrangement), the "E" comes immediately after the "TH", so there are only 81 ways of positioning the "TH" and the "E" that meet both constraints. But then you have to arrange the other letters as well.

karkar athlete

hivizman thanks a million , I searched every book to do with algebra with contained permutations and combinations in the college library today:eek: for a similar example but there was none with such restraints as in the question I was given.

I see now that I had considered all constraints with the question to get 81. I just can’t get my head around permutations and combinations and unfortunately I did not do this on the Leaving Cert course either.

But thanks once again , I have spent ages looking at the question trying to do it but gave up numerous times as I just got frustrated.

I am sorry about this but there is another question I am unsure about which is "How many numbers greater than 3,000,000 can be formed by arrangements of 1,2, 2,4,6,6,7?"

I am right to say that I can assume we are not allowed repetitions of the numbers, but even with that I tried multiplying out some of the numbers and the biggest I got was 4,000 odd, so is there actually combinations of those numbers which is >3,000,000.

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henbane

karkar athlete wrote: »

I am sorry about this but there is another question I am unsure about which is "How many numbers greater than 3,000,000 can be formed by arrangements of 1,2, 2,4,6,6,7?"

I am right to say that I can assume we are not allowed repetitions of the numbers, but even with that I tried multiplying out some of the numbers and the biggest I got was 4,000 odd, so is there actually combinations of those numbers which is >3,000,000.

Similar to the MATHEMATICS questions, look for all numbers with leading 4, 6 or 7 as per hivizman's method above e.g. all permutations of the the other 6 numbers (minus duplicates) * 3