How many words of $13$ letters can be constructed from the English alphabet which contain $4$ vowels and $9$ different consonants (vowels can be the same). This is what I think: Pick the four vowels (does this arrange them as well?) = $5^4$. Choose the consonants = $\binom{13}9$ Now then arrange these consonants in $9!$ ways Total = $5^4 \binom{13}9 9!$.    My friend got this: $5^4 \binom{21}9 \binom{13}4$ to arrange the four vowels (with the last binomial).
122418
How many words of $13$ letters of the English Alphabet can we constructed which contain $4$ vowels and $9$ different consonants?    This is what I did: Choose the consonants (diff.) in $\binom{21}9$ ways. Arrange these consonants in $9!$ ways. Now we have $4$ empty spots remaining for the vowels, which can be chosen in $5^4$ ways.    My question is: Is this the correct way to think about it? I'm unsure whether the $5^4$ arranges the vowels or not,
91071
In how many ways can the letters of the word CHROMATIC be arranged,      a.  without any restrictions  b.  if the position of the vowels can’t change  c.  find the probability that the string of letters begins with the letter M
50204
I'm not sure what these are called, but how can I form a word like "quadruple" for any number I want? Like 5× as much is quintuple, what is 31× as much or 147× as much? I want to know how they are formed so I can make my own.  Similarly, how do I construct the prefixes, such as unicycle, bicycle, tricycle, quadricycle? I figured out 12 is duodeci-, but how can I form any prefix like this?
46854
I need to calculate the possible combinations for 8 characters password. The password must contain at least one of the following: (lower case letters, upper case letters, digits, punctuations, special characters).  Assume I have 95 ascii characters (lower case letters, upper case letters, digits, punctuations, special characters).   lower case letters = $26$ upper case letters = $26$ digits = $10$ punctuations &amp; special characters = $33$   The general formula for the possible passwords that I can from from these 95 characters is: $95^8$.  But, accurately, I feel the above formula is incorrect. Please, correct me. The password policy requires at least one of the listed above ascii characters. Therefore, the password possible combinations = $(26)*(26)*(10)*(33)*(95)*(95)*(95)*(95)$  Which calculation is correct?  EDIT: Please, note that I mean 8 characters password and exactly 8. Also, There is no order specified (i.e. it could start with small letter, symbol, etc.). But it should contain at least one of the specified characters set (upper case, lower case, symbol, no., etc.).
18980
Is there a general formula for the number of distinguishable arrangements of length $\ell$ of a sequence of size $s$, given the repeated elements? So,  For example, would there be a general formula to solve the problem:  How many distinguishable 5-letter arrangements are there of the word "PROBABILITY" ?  I'm looking for a general formula.
36007
If from a deck of 52 cards, I extract 10. In how many combinations do you get at least one ace?  I have come up with two possible answers, but I don't know which one is the right one and why.  So one is $$4{51 \choose 9}$$  the reasoning is that first I extract an ace from the four that there are, and then I have ${51 \choose 9}$ combinations for the other 9 cards.  The second is  $${52 \choose 10} - {48 \choose 10}$$  reasoning that there are ${52 \choose 10}$ total combinations of ten cards and I subtract ${48 \choose 10}$ combinations without any aces.  So from other questions it seems that the first one is the correct answer, but why would the second one be wrong?  Thanks for the answers.
115288
To find the number of three letter words that can be formed from the word 'SERIES', with or without meaning and without repetition.   The number of permutations if all letters were distinct  = $^{6}P_3$.    As 'S' and 'E' are repeating, the arrangement in which 'S' are interchanged are the same, so we divide by $2!$ for 'S' and likewise for 'E'.    Number of permutations $= \frac{^{6}P_3}{2!.2!} = 30$   Which isn't the correct answer. I would like to get some help here.
58396
I have a large dictionary file with 300,000+ words in it and I'm trying to find all words with the vowels aeiou in that order and have only exactly 5 vowels. My current attempt does not seem to be working and for the life of me I don't understand why.  less mywords | grep -iE [^aeiou]*a[^aeiou]*e[^aeiou]*i[^aeiou]*o[^aeiou]*u[^aeiou]*   I think it get's all the words I'm looking for but there are a few words that pop up which I can not get rid of.  abstemiousnesses ultraserious   There are a few other but they are in the same vein. Curiously, even if I add something like [^u]* to the front ultraserious keeping popping back up! Any solution would be fine however I would like it restricted to grep as we've been told it can be done using only grep.
52465
Given the letters A, A, B, B, C, D, E   How many different arrangements are possible that begins and ends with the letter "A"?    B,B,B,G,G  Three boys and two girls shown above are lined up side by side in a single row so that boys and girls are lined up in groups respectively. How many different arrangements of the childern are possible?  There are 3 Republicans and 2 Democrats on a Senate committee. if a 3-person subcommittee is to be formed from this committee, what is the probability of selecting 2 Republicans and 1 democrat?  An urn contains 4 white marbles and 5 black marbles, all of equal size. If two marbles are drawn at random with no replacement, what is the probability that 2 marbles are different colors?  If the 5 cards A, B, C, D, E are placed in a row so that cards B and C must be next to each other, how many different arrangements are possible?
5024
Problem: Find the number of twelve-card hands containing four cards in each of three suits.   My solution:   pick a suit : 4 ways. choose 4 cards from the 13 in that suit: ${^{13}}{C}_{4}$ ways. pick another suit: 3 ways. choose 4 cards from the 13 in the new suit: ${^{13}}{C}_{4}$ ways. pick the final suit: 2 ways. choose 4 cards from the 13 in the final suit: ${^{13}}{C}_{4}$ ways.   Hence, total number of ways is $4\cdot{^{13}}{C}_{4}\cdot3\cdot{^{13}}{C}_{4}\cdot2\cdot{^{13}}{C}_{4} = {^{4}}{P}_{3}\cdot\left({^{13}}{C}_{4}\right)^3 $ total hands.   This is the incorrect answer. Why?
2928
If the all S,P are separated in the word MISSISSIPPI then the total possible arrangements are $k\frac {10!}{4!.4!} $ then value of k is? Now all S,P are separated so we can have slash method. Thus they can be placed alternatively in $6! $ ways and rest 5 can be placed in $5! $ ways. Thus we equate to get $$6!.5!=k\frac {10!}{4!.4!} $$  to get $k=13.7$ while $k=4/3$ .Where is my mistake Thanks
59422
In school we are taught the vowels: A, E, I, O, U, and sometimes Y.    got me thinking about when the letter Y is considered to be a vowel.  I understand (perhaps incorrectly) that in words like bicycle and why it is a vowel. What about the word voyeur (as mentioned in the XKCD alt-text)?   If I've got this backwards, and Y is almost always a vowel, how can I tell when it is a consonant? Thinking back, I don't think my education ever covered the difference between them, we just memorized which letters were which.
54578
It's Friday again, how about some fun to get us into the weekend?  What is the longest word you can come up with for which all the letters in that word are in alphabetical order?   Rules:   English words only    Can't be a name of a place, person or other proper noun.   if it contains the same letter twice in a row, that does not disqualify it.   No fair looking it up on Google!   Update: One more rule to help you guys out.   The word can be in either ascending or descending alphabetical order.
63537
Given an alphabet of size $n$, how many strings of length $c$ contain every single letter of the alphabet at least once?  I first attempted to use a recurrence relation to work it out:  $$ T(c) = \left\{ \begin{array}{cr} 0 &amp;\mbox{ if $c&lt;n$} \\ n! &amp;\mbox{ if $c = n$} \\ T(c-1) \cdot n \cdot c &amp;\mbox{ if $c &gt; n$} \end{array} \right. $$  As there's no strings that contain every letter if c &lt; n, and if c = n then it's just all permutations. When c > n you can take any string of size (c-1) that contains all letters (of which there are $T(c-1)$ to choose from), you choose which letter to add (of which there are $n$ choices) and there are $c$ different positions to put it. However, this gives out results that are larger than $n^c$ (the total number of strings), so it can't be right, and I realised it was because you could count some strings multiple times, as you can make them taking different inserting steps.   Then I thought about being simpler: you choose n positions in the string, put each letter of the alphabet in one of those positions, then let the rest of the string be anything:  $$ {c\choose{n}} \cdot n! \cdot n^{c-n} $$  But again this counts strings multiple times.  I've also considered using multinomial coefficients, but as we don't know how many times each letter appears in the string it seems unlikely they would be much help. I've also tried several other methods, some complicated and some simple, but none of them seem to work.  How would you go about working out a formula for this? I'm sure there's something simple that I'm missing.
51585
How many strings with seven or more characters can be formed from the letters in EVERGREEN if the two R's are consecutive?  I know how to do this if the two R's don't have to be consecutive. Please point me in the right direction.
11252
I know this problem is familiar to "the number of binary strings of length $n$ with no consecutive zeros" but still it is a way different problem. But in that question the alphabet is set to $\{0,1\}$ and here:  Q:  Let $A=\{0,1,2\}$ be an alphabet, what is the the number of possible strings of length $n$ with no consecutive zeros in it?
86748
I had a question I was hoping for some help on:     Find all of the subgroups of $A_4$   Here is what I know:  $A_4$ is the alternating group on 4 letters. That is it is the set of all even permutations. The elements are:  $(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)$   which totals to 12 elements. Which means, the subgroups should have order 1,2,3,4,6 and 12. The groups of order 1 and order 12 are trivial. I also know the groups of order 2 which are:  $[1,(12)(34)], [1, (13)(24)], [1, (14)(23)]$  but would someone be able to help me find the subgroups of order 3, 4 and 6 and be able to provide a good, easy explanation of how you got to them for me? I'm pretty sure there is no such groups of order 6, but I don't know how to explain why.  Thank you so much for your help in advance, I really appreciate it!