I know a lot of similar questions have been asked here, but sadly I can't seem to wrap my head around this topic. The topic is combinatorics.
I am trying to get into math during high school to be more ready for college and this is so hard.
I have a problem.
Making permutation from word PAPARAZZI.
- I need to find all possible permutations of this word which was $\frac{N!}{N_1!N_2!\ldots}$ where $N$ is the number of characters and $N_1$ etc. are groups of duplicates. So for this example it was $\frac{9!}{3!2!2!}$.
Trouble comes with the second and third problems.
- How many permutations can I make if there always has to be exactly $2$ characters between my Zs?
- How many permutations can I make if P and I cannot be next to each other?
I thought that for the "two chars problem" I could take the Zs as single block and just do $\frac{8!}{3!2!}$ and for the second one Reduce total number of chars to $6$ ($9$ – both Ps and I) and continue with $\frac{6!}{3!2!}$, but sadly both answers were wrong and I haven't got any feedback because of distance education, so I wanted to ask what I did wrong or what I do not understand?
Thanks in advance.
Best Answer
Your answer to the first question is correct.
Since there must be two characters between the two Zs and there are nine letters in the word PAPPARAZZI, the first Z must appear in one of the first six positions. Once we choose the positions of the two Zs, we have seven letters to arrange in the remaining seven positions, including three As, two Ps, one R, and one I. Choose three of these seven positions for the As, two of the remaining five positions for the Ps, and one of the two remaining positions for the R. The I must go in the final open position. Hence, there are $$\binom{6}{1}\binom{7}{3}\binom{4}{2}\binom{2}{1} = 6 \cdot \frac{7!}{3!2!1!1!}$$ admissible permutations.
As you determined, there are $$\binom{9}{3}\binom{6}{2}\binom{4}{2}2! = \frac{9!}{3!2!2!1!1!}$$ distinguishable permutations of the letters of the word PAPPARAZZI.
From these, we wish to subtract those permutations in which P and I are adjacent.
P and I are adjacent: Then we have eight objects to arrange: three As, two Zs, a P, an R, and a block consisting of an adjacent I and P. Choose three of the eight positions for the As, two of the remaining five positions for the Zs, then arrange the three distinct objects in the remaining three positions. The I and P can be arranged within the block in $2!$ ways. Hence, the number of arrangements in which P and I are adjacent is $$\binom{8}{3}\binom{5}{2}3!2! = \frac{8!}{3!}$$
However, if we subtract this amount from the total, we will have subtracted arrangements in which both Ps are adjacent to I twice, once when we subtracted arrangements containing PI and once when we subtracted arrangements containing IP. We only want to subtract them once, so we need to add those arrangements to the total.
Arrangements in which both Ps are adjacent to I: We have seven objects to arrange: three As, two Zs, R, and the block PIP. Choose three of the seven positions for the As, two of the remaining five positions for the Zs, and one of the remaining two positions for the R. The block must occupy the remaining position. Hence, there are $$\binom{7}{3}\binom{4}{2}\binom{2}{1} = \frac{7!}{3!2!1!1!}$$ such arrangements.
By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\frac{9!}{3!2!2!1!1!} - \frac{8!}{3!} + \frac{7!}{3!2!1!1!}$$