Reputation
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
1 15 40
Newest
 Necromancer
Impact
~98k people reached

May
3
answered How can I prove using induction that the Hadamard matrices are orthogonal?
Apr
29
comment How many $A_5$ are there inside $A_6$?
...the labelings that correspond to nonzero powers of the cycles $(23456)$ and $(23546)$ but these all give the same two copies of $A_5$. Similar considerations imply that each of the $45$ order-$2$ permutations appear in exactly two copies of $A_5$ of the second type.
Apr
29
comment How many $A_5$ are there inside $A_6$?
@monomorphic: in the copies of $A_5$ that relate to the rotational symmetries of the icosahedron, which $15$ of the $45$ order-$2$ permutations appear is determined by the labeling of the vertices. For example, if the north pole is labeled $1$ and the adjacent vertices are labeled $2$ through $6$ in clockwise order, then the rotation about the center of the edge joining $1$ to $2$ corresponds to the permutation $(12)(36)$. This permutation will also appear when the labeling of the vertices adjacent to $1$ is $2$, $3$, $5$, $4$, $6$ in clockwise order. It will also appear in...
Apr
29
comment How many $A_5$ are there inside $A_6$?
@monomorphic: in the copies of $A_5$ in $A_6$ that leave one element unmoved, all $15$ permutations of order $2$ involving the other five elements appear. Hence each permutation of order $2$ appears in two of the six copies of $A_5$ of the first type, namely the two that omit either of the two elements fixed by the permutation (and hence contain all four elements moved by the permutation).
Apr
29
comment How many $A_5$ are there inside $A_6$?
@monomorphic: actually each subgroup of $A_6$ of order $2$ lies in two of the six copies of $A_5$ of the first type and in two of the six copies of $A_5$ of the second type. A subgroup of $A_6$ of order $2$ can only consist of the identity and a permutation that is the product of two $2$-cycles. Such a permutation moves four elements and fixes two. There are $15$ such permutations in $A_5$ and $45$ such permutations in $A_6$.
Apr
28
comment How many $A_5$ are there inside $A_6$?
@monomorphic: Can you elaborate on your question? Are you asking whether every subgroup of $A_6$ of order $2$ is a subgroup of an $A_5$ of at least one of the two classes? ...of exactly one of the two classes?
Apr
26
answered Steiner Triple System
Apr
26
revised Steiner Triple System
added 222 characters in body
Apr
26
comment Steiner Triple System
My original derivation of the statement that the number of pairs not containing $a$ is $\frac{n-1}{2}$ was flawed. I've corrected this now. Sorry for all the mess. You really should focus on understanding Alexey Burdin's answer rather than mine, but I leave this here for the time being since you've already spent some time on it. In partial recompense, I offer a completely different approach, not using inclusion-exclusion, as a separate answer.
Apr
26
comment Steiner Triple System
Much more straightforward to do things Alexey Burdin's way. I should have defined $T_s$ to be the set of triples that DO contain the elements of $s$. With this definition of $T_s$ you want to compute $$\lvert(T_{\{a\}}\cup T_{\{b\}}\cup T_{\{c\}})'\rvert=\lvert T\rvert-\lvert T_{\{a\}}\rvert-\ldots+\lvert T_{\{a,b\}}\rvert+\ldots-\lvert T_{\{a,b,c\}}\rvert.$$ This leads directly to Alexey Burdin's much simpler answer.
Apr
26
comment Steiner Triple System
Sorry for the confusion! I got off on the wrong foot with my explanation by defining $T_s$ as the set of triples NOT containing the elements of $s$. I think this made things hopelessly confusing, essentially because it required thinking in terms of double negatives, and because it required an unusual form of inclusion-exclusion (intersections on the left, unions on the right, rather than the reverse). It also required a bit more algebra than necessary.
Apr
26
comment Steiner Triple System
@AlexeyBurdin: you should post your comment as an answer. I managed to tie myself in knots with the inclusion-exclusion argument. Yours is much simpler.
Apr
26
revised Steiner Triple System
added 26 characters in body
Apr
26
revised Steiner Triple System
added 26 characters in body
Apr
26
revised Steiner Triple System
added 26 characters in body
Apr
26
answered Steiner Triple System
Apr
17
comment How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels?
The problem is asking for the union of the set of sequences such that the first letter is not a vowel with the set of sequences such that the last letter is not a vowel. This is consistent with the two stated solutions (numerical value 160665120) and with the phrase "possibly both", which suggests that it is not necessary that both letters not be vowels.
Apr
17
comment How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels?
If you do that, then you will have double counted the case "neither first nor last letter a vowel". This would then have to be subtracted away to get the correct count. Better not to include it in the first place. Of course it's even better to go with one of the methods that involves fewer cases.
Apr
16
comment How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels?
...have the middle four letters in VVVC (V=vowel, C=consonant) order, but there are, in fact, four possible orders: VVVC, VVCV, VCVV, CVVV. In general, if the middle four letters include $r$ consonants and $4-r$ vowels, there are $\binom{4}{r}$ orders. If you include these binomial coefficient factors, your answer will agree with the stated solutions.
Apr
16
comment How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels?
Question: instead of "last letter not a vowel", did you mean "last letter not a vowel but first letter is a vowel"? I think you need the extra stipulation to avoid overlap with the "first letter not a vowel" case; the extra stipulation is also consistent with your not having included "0 vowels are used previously" among your cases. If that's what you meant, then your analysis was close to being complete. You omitted one thing: in the "4 vowels used previously" case, you...