Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question?
The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E,
(b) the letters I and P are next to one another,
(c) there are three letters between N and T,
(d) there are at least three letters between N and T.
EDIT: I didn't spot the duplicate R, that will change things substantially! Hey there Tim, welcome!!
super happy to have you around the forums
I can definitely help you out!
Okay, so for all of these questions, we will need the total number of arrangements of the letters. The number of ways we can arrange unique objects in a line is just n!, where n is the number of objects. However, we have two duplicate letters,
so we must divide that by 2! to compensate:Okay, for Part A, we just put the E down first, then arrange the rest of the letters in a line behind it. So, we are only arranging 6 elements, meaning, 6! is the number of ways we can do that. We put that over the the total number of arrangements to get the answer.
We can ignore the 2! here, because it would be present on both the top and bottom, and would thus cancel.For Part B (and this is a common approach), let's consider the P and I as a single letter grouped together (taped together, can't be pulled apart). So, we are now arranging 6 elements again, since the P is stuck to the I. However, we can THEN swap the P and I around, swap the order they are taped. This gives the calculation below, 6! multiplied by 2 for the swap.
Again, we can ignore the 2! in this calculation:Three letters between N and T requires a similar approach, but NOW we bring in permutations. First, we need to select the letters to put between the N and the T, and the order counts, so we use a permutation (forgive my terrible version of the permutation notation below).
However!! The number of ways we can arrange these letters depends on whether we get 0, 1 or 2 R's (this is a nasty question).
If we have 0 R's, there is only we are selecting 3 from the 3 non R letters:
Then, we swap the N and T as before, then consider the ordering:
If we have 1 R, then we have selected 2 non R letters to go with it. So, we use a combination to make those 2 selections, then multiply by 3! to arrange:
We then multiply by two and arrange, no duplicates in the arrangement:
If we have two R's, we've picked one non R letter (3 ways to do this), then we arrange it with 2 R's. Similar to above, this means dividing the final result by 2! because we have two alike elements.
Then, we arrange, no duplicate elements in the final arrangement either:
And now, we add these to get the final probability (same answer as before, did I screw up?)
I'm going to leave the final part for you, but basically, we want
at least 3, so you'll be applying the principles I used above for 4 and 5 letters, then adding together (there are other methods as well).
- Find the number of ways we can arrange the letters between N and T
- Double this result because we can swap N and T around
- Arrange the grouping as a BIG LETTER using the factorial principle
I hope this helps! Let me know if you need anything explained in a little more detail, I moved fast
I tried to take my working already and adapt it to the duplicate letter, I'm not 100% confident, if someone spots an error please correct me!