2 small permutation and combination questions

[Ttwnycorporation]

1. Find the number of different permutations of the word "ASSESSMENT" if the first and last places are occupied by consonants.

For 1, I think the complementary cases method is supposed to be used, but I'm clueless

2. Find the total number of 3-letter code words that can be formed using the word "HAPPY".

For 2, which is more of a concept question, why is it that when I consider the case when 2 Ps are inside the code word, the number of code words I can get = 3!/2!*3? Shouldn't it be 3!/2! ways, since I'm permuting 3 letters with 2 similar letters?

Thank you in advance, and I wish you all a good day!!! EDIT: THERE IS NO REPETITON OF LETTERS IN THESE 2 QEUSTIONS GUYS! IMPORTANT!!! OK GOODBYE

 

[Dr.Peterson]

For (1), I would consider the cases where both first and last are S, where one of those is S, and where neither is S.

For (2), when both P's are used, there are 3!/2! = 3 ways to place the P's, and 3 ways to choose the other letter.

 

[HallsofIvy]

1. Find the number of different permutations of the word "ASSESSMENT" if the first and last places are occupied by consonants.

For 1, I think the complementary cases method is supposed to be used, but I'm clueless

There are only four different consonants, S, M, N, and T so "first and last places are occupied by consonants" must be
S__S
S__M
S__N
S__T
M__S
M__N
M__T
N__S
N__M
N__T
T__S
T__M
T__N
a total of 4+ 3+ 3+ 3= 13 possibilities. There are, in each case 8 letters left which have 8! combinations. There are a total of 13(8!) such combinations.

2. Find the total number of 3-letter code words that can be formed using the word "HAPPY".

For 2, which is more of a concept question, why is it that when I consider the case when 2 Ps are inside the code word, the number of code words I can get = 3!/2!*3? Shouldn't it be 3!/2! ways, since I'm permuting 3 letters with 2 similar letters?

Thank you in advance, and I wish you all a good day!!! EDIT: THERE IS NO REPETITON OF LETTERS IN THESE 2 QEUSTIONS GUYS! IMPORTANT!!! OK GOODBYE

Do you mean "APP" would not be valid for (2)? Since "permutations of a word" includes all letters in the word, I don't see how this would apply to (1).

 

[Ttwnycorporation]

Sorry, but won't there be extra cases for your solution to number 1, Since the remaining letters contain 2 Es? And there are a few cases where there will be 2S and 2E in the eight letters, so don't I have to divide them as well?
For question 2, I was asking for how to find the total number of ways you could arrange the word with 2Ps inside, it's another question entirely. Thank you so much for the help!!

 

[pka]

Sorry, but won't there be extra cases for your solution to number 1, Since the remaining letters contain 2 Es? And there are a few cases where there will be 2S and 2E in the eight letters, so don't I have to divide them as well?
For question 2, I was asking for how to find the total number of ways you could arrange the word with 2Ps inside, it's another question entirely. Thank you so much for the help!!

Prof. Ivy correctly identified the thirteen possible cases. \(S\_\_S\) The remaining eight letters can be arranged in \(\dfrac{8!}{(2!)^2}\) ways.
There are six cases that either began with \(S\_\_X\) or \(X\_\_S\) end with an \(S\) NOT BOTH. The remaining eight letters can be arranged in \(\dfrac{8!}{(2!)(3!)}\) ways
There are six cases the neither begin with nor end with an \(S\). The remaining eight letters can be arranged in \(\dfrac{8!}{(2!)(4!)}\) ways.
Now that count is \(\dfrac{8!}{(2!)^2}+6\cdot\left(\dfrac{8!}{(2!)(3!)}\right)+6\cdot\left(\dfrac{8!}{(2!)(4!)}\right)\).

2. Find the total number of 3-letter code words that can be formed using the word "HAPPY". I would have done the differently than the other two profs. Two cases: one includes two \(P's\) the other has three distinct letters.
The first count is \(3\cdot\dfrac{3!}{2!}\). The second count is a simple permutation, \(\mathcal{P}_3^4\) That is \(9+24\)

 

[Dr.Peterson]

Sorry, but won't there be extra cases for your solution to number 1, Since the remaining letters contain 2 Es? And there are a few cases where there will be 2S and 2E in the eight letters, so don't I have to divide them as well?
For question 2, I was asking for how to find the total number of ways you could arrange the word with 2Ps inside, it's another question entirely. Thank you so much for the help!!

I'm not sure which of us you were responding to. But each case can be treated differently, and in my case there may be cases within a case. There's a good bit more to do after each of our divisions into cases. I lumped cases together more, which I think is easier, but that leaves more splitting to be done.

We'll need to see how you applied one of our suggestions, in order to be sure you've got it.