How many arrangements are possible with five letters chosen from Mississippi? How would I approach this problem? I just want some hints.
3292
How many 6-letter permutations can be formed using only the letters of the word, MISSISSIPPI? I understand the trivial case where there are no repeating letters in the word (for arranging smaller words) but I stumble when this isn't satisfied. I'm wondering if there's a simpler way to computing all the possible permutations, as an alternative to manually calculating all the possible 6-letter combinations and summing the corresponding permutations of each. In addition, is there a method to generalize the result based on any P number of letters in a set with repeating letters, to calculate the number of n-letter permutations?  Sorry if this question (or similar) has been asked before, but if it has, please link me to the relevant page. I also apologize if the explanation is unclear, or if I've incorrectly used any terminology (I'm only a high-school student.) If so, please comment. :-)
81716
I'm at the part where I am asked to complete a sequence of 7 letters and a question mark. I simply just don't understand what I am supposed to do. Can I get some help?
89450
I'm stuck with the basic counting problem.  Problem is the following:     Corrupt professor Z has a class of 50 students. He needs to give exactly 10 A's. However five students already have a special deal (they are professor Z's nephews and neices) and will get A's for sure. How many ways can the 10 A's be distributed?   My thought : 45 Choose 5, because we can ignore 5 people(nephews and nieces) who will get A's grade for sure.   I'm not sure my thought is correct or not. I'm appreciated any comment or explanation for this.
103156
I have no idea what the first problem on this list is asking? The others I can figure our, but does anyone know what the first one is actually asking? If so, how would you set it up?  Nate can only invite 8 people to his birthday, but has 13 friends he’d like to invite.  Eight of his friends are boys and five are girls.  Three of the girls are roommates and must stay together.  If Dante randomly selects 8 people, determine the probability for each scenario.   Dante’s random selection is a valid outcome Dante does not invite any girls Dante invites an equal number of boys and girls
86551
How many arrangements are there of MATHEMATICS with both T's before both A's or both A's before both M's or both M's before the E ?  Can someone also point to some online resource that has such practice questions ?  Thanks.  Edit:  I have some initial thought, please tell me if I am in the right direction :  For the case both T's before both A's :  We can picture the permutation as follows :  (7) T (7) T (7) A (7) A (7)  (Remaining letters: M H E M I C S - total 7)  Each of the brackets are empty spaces for now. We need to pick 7 out of the total 7 X 5 = 35 spaces; that will ensure both T's being before both A's.  So number of ways = C(35,7) X 7!/2! (because two M's are same - we need to divide by 2!)
116996
What is the number of ways we can choose three letters from $\{A,B,...,Z\}$ if repetition is allowed?      This is not a permutation.  Is there any formula to solve this kind of problems?
112900
How many different/unique $4$-letter arrangements are there of the letters in the word Mississauga?  I'm thinking we need permutations? I already tried finding each letter and still don't get it... any help?   So far I've broken it off to:-  $M=1$  $I=2$  $S=4$  $A=2$  $U=1$  $G=1$  However, I don't really know what to do next.
106910
Im using the principle on inclusion and exclusion to solve this  There are 4 cases C1,C2,C3 ,C4 respectively   So taking the case where CAR DOG and BYTE comes together  n(C1^C2^C4)=20!   But in my book its given as 19!. This is my explanation Taking CAR,DOG AND BYTE together total 10 letters so (26-10)!=16!  These letters are considered as one word so (16+1)!=17!  But there will be 3 inter arrangements among them so (17+3)!=20!   Where am i doing wrong?   EDIT: You need to find 26!-N(C1UC2UC3UC4) For this i need to calculate each permutations involving the words so my question is regarding a particular case
50261
We are provided with a string A (may contain repeated characters as well ). One needs to make all possible permutations of A , and find out how many pairs of strings chosen from these permutations are not similar.  Two strings are similar if we can make both of them exactly same by carrying out at most 1 swap in each of them .  Example :- if we are provided with string abcd  , then for permutation abcd , the permutations which are not similar to it are bcda , bdac , cadb , cdba , dabc , dcab . Similar is the case for other permuations as well. So overall number of desired pairs are 24*6=144 .  I tried solving it for each possible pair by approach mentioned here     But as the permutations can be large enough , I need a better approach of solving this.
18379
How many 5-cards poker hands are there containing at least 3 of the 4 suits?  I don't how know how to properly handle this question.  There 13 ranks of each card, and 4 suits  My Attempt:  Begin with the first 3: C(13,3) X 4 ways to choose suits  Then pick last 3 cards of anything:  C(49,2)  My solution: 1,345,344  Thanks,
91440
How many words can be formed from the word 'alpha'? The letter 'a' may be used twice but the other letters may only be used once.  There are no restrictions on whether or not they're real words, just combinations of letters. So, I need to add together:  $5$ letter words $+$ $4$ letter words $+$ $3$ letter words $+$ $2$ letter words $+$ $1$ letter words  There are, $(5!/2!)=60$, 5 letter words. 120 total but then we divide out by duplicate words because we have two a's.  $1$ letter words $=4$  The part I am having trouble with is the duplication of 'a'. I can't seem to figure out how to find $2,3,$ and $4$ letter words. For 2 letter words, there were so few examples I just wrote them out and saw that there were 13. So,  $(5!/3!)-[(2!)3+1]=13$  20 possibilities and then we need to subtract out the ones that contain 2 of p,l, and h, as well as the second aa.  I'm having a hard time believing my work for 2 letter words is accurate and attempting to do 3 and 4 letter words suggests that it isn't. Let me know if there is anything I can to do clear up what is being asked. Thanks in advanced.
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,
112368
Say I have 4 pairs of identical pens (say red, blue, green and black).   How many ways can I arrange them such that no two identical pens are next to each other?  Inclusion/Exclusion works (I get 864) but it seems a bit long. Is there another way of doing this problem?  I noticed that 864 is $4 \times 6^3$ or in other words $4 \times \binom{4}{2}^3$, perhaps this could be handy?
91895
How many words can be formed from the letters of the word 'DAUGHTER' so that the vowels never come together ?  The answer is obviously $8!-6!\cdot3!$.  My question is that if we ponder from a different perspective, that is taking $5$ consonants first and arranging them ($5!$ ways of doing that) and then placing the $3$ vowels in the $6$ places created due to the arrangement of consonants ($\frac{6!}{3!}$ ways to do that), the answer should be $5!\frac{6!}{3!}$.  What is wrong with this?
79254
I have to figure out a way to count how many sequences are there s.t:   My alphabet is $1$ up to $49$. Each number that is chosen, is chosen only once. The sequence is $6$ digits long. The sequence is in increasing order.  e.g that is a valid sequence: $(1, 2, 40, 42, 43, 49)$  and this is not: $(1, 2, 40, 25, 25, 12)$   I know that for the first 3 rules the answer is: $\binom{49}{6}\cdot 6!$  How do I deal with the fourth?
7424
In how many ways can we choose k items from n distinct items put in a circle.   Then further extension of the question, if I want to find out the number of ways to choose k items from n distinct items in a circle such that no 2 are together.  Thanks
80510
How many 4 digit numbers have a sum of 9?  I converted that into: How many ways can you rearrange 3 stars and 9 bars?  But the first one has to be a number so it's actually: How many ways can you rearrange 3 stars and 8 bars?  I can convert it but I forgot how to solve the stars and bars problem? Can someone explain how to solve a converted stars and bars problem without giving me the answer to this question? Thanks.