Feb26 revised Prove that a k-regular connected graph with edge chromatic number = k is 2-connected. dollar signs added where needed Feb26 suggested suggested edit on Prove that a k-regular connected graph with edge chromatic number = k is 2-connected. Feb20 revised Homomorphism space of permutation modules added 4 characters in body Feb19 comment Homomorphism space of permutation modules Yes, I know that. I was only wondering if there is a way to find $H$ without having to do some (often messy) calculations (to find such $H$). Thanks anyway! Feb19 accepted Homomorphism space of permutation modules Feb19 comment Homomorphism space of permutation modules Thank you. Is there a "constructive way" to find the subgroup $H$ in the above? Also, isn't the number of "H"-orbits in $Y$ equal to the number of $G$-orbits of $X \times Y$? Feb19 revised finding how many passwords can be made using an arrangement of letters added 102 characters in body Feb19 answered finding how many passwords can be made using an arrangement of letters Feb19 revised Homomorphism space of permutation modules edited title Feb19 revised Homomorphism space of permutation modules added 122 characters in body Feb19 asked Homomorphism space of permutation modules Feb17 comment $A=(A \cap B) \cup(A \cap B^\mathsf{c})$ Or by distributive law $(A \cap B) \cup(A \cap B^\mathsf{c})=A \cap (B \cup B^\mathsf{c})=A$ Feb10 comment Find $\log_{7}\sqrt[3]{xy}$ in terms of $p$ and $q$. Hints: $\log(ab)=\log(a)+\log(b), \log(a^b)=b\log(a)$ Feb2 comment Number of Contiguous Arrangements of Four Books out of Twelve @Newb it is just a matter of "estimation". You have 12! which is about 400 million possible arrangements of the books and 192 seems unrealistic, given that we have only a constraint on 4 of the books! Feb2 comment Number of Contiguous Arrangements of Four Books out of Twelve @Newb also, i believe you need to permute the 8 remaining books as well. So the answer should be 9*8!*4!. Feb2 comment Number of Contiguous Arrangements of Four Books out of Twelve @Newb another way of saying this: put the non-blue 8 books in a row, now it is just a matter of choosing where to put the 4 books (thought as 1 book). We can put the 4 books before all 8 non-blue books, or between the first and the second of the non-blu books, etc etc but you can also put them aftell ALL 8 non-blue books. Feb2 comment Number of Contiguous Arrangements of Four Books out of Twelve @Newb I might be very tired but you have four blue books, you are left with 8 books. Now let an S indicate each of the remaining 8 books and - be a space (where to put all the 4 books!), then we have this configuration: -S-S-S-S-S-S-S-S- . We have 9 bars. Feb2 comment Number of Contiguous Arrangements of Four Books out of Twelve I think there are 9 places where to put the 4 books (just think of them as 1 book to place among 8 books). Feb1 revised Is the modulo operation a standard mathematical operation which actually means Euclidean division? inappropriate tags Feb1 suggested suggested edit on Is the modulo operation a standard mathematical operation which actually means Euclidean division?