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4

HINT: there are $6^6$ such numbers. One-sixth of them have 1 as their ones digit, and each one of them contributes 1 to the sum. One-sixth of them have 2 as their ones digits, and each one of them contributes 2 to the sum. And so forth. Similarly, one-sixth of these numbers have 1 as their tens digit, and each one contributes 10 to the sum. One-sixth ...

4

For each element of $A$ you have $k$ choices, so there are $k^n$ possibilities. Given $A$, for each element of $B$ you have $k-1$ choices, so there are $(k-1)^n$ possibilities. There are therefore $k^n(k-1)^n$ pairs possible.

3

The group you describe is indeed a subgroup of $S_n$, and it is isomorphic to $S_a\times S_b$. The symmetric group $S_n$ acts naturally on the set $\{1,\ldots,n\}$ and the corresponding subgroups $S_a$ and $S_b$ act naturally on the sets $\{1,\ldots,a\}$ and $\{a+1,\ldots,n\}$. These sets are the orbits of the action of $S_a\times S_b$ on the set $\{1,\ldots,... 3 I can tell from a general familiarity with these types of problems that$n$refers to the combined number of$0$'s and$1$'s. If they omitted the restriction that$1$s occur in groups of three or more, there would be$2^n$arrangements. However, you should double check this with whoever gave you the problem. Also, the all zeroes sequence should be ... 2 Try to count the opposite: The number of permutations with A not in the first place AND with E in the last place. There are 6! permutations with E in the last place. From the remaining 6 letters there are 5! with A in the first place (all permutations of RTICL). So we have 6!-5! permutations with A not in the first place and E in the last place. The total ... 2 If it is known or not I don't know, but for the latter part: Start by noticing that for any valid permutation$\sigma \in F(n,k)$you have for the largest number$n$the condition$\sigma(n) > n - k$and thus exactly$k$possibilities for$\sigma(n)$. For the next biggest number$n-1$we have the condition$\sigma(n-1) > n-k-1$so the ... 2 This statistic is in the FindStat database, you find it and code to produce it at https://www.findstat.org/St000019. The next row is 1, 9, 52, 252, 1146, 5226, 24892, 125316, 642581, 2829325 What you computed is, up to a relabelling of the numbers$1$through$n-1\$, the same as the connectivity set: First observe that your definition depends on the ...

2

Here, I will use the following facts: $$C^n_{n+1}=C^n_{-1}=0\tag{1}$$ $$C^{n+1}_i=C^n_{i-1}+C^n_i\tag{2}$$ I will also be assuming some basic knowledge of re-indexing sums, and assuming that you meant to say $$B = \sum^{n+1}_{i=0}C^{n+1}_i (X_iX_{n-i+1}),$$ rather than $$B = \sum^{n+1}_iC^{n+1}_i (X_iX_{n-i+1}).$$ With that in mind, we see that $$\begin{... 2 Not necessarily (False); you can have 5 independent vectors in \Bbb R^6. A singular matrix has determinant 0, so A^TA also has determinant 0 and is singular. (True) A permutation matrix determinant is (-1)^n, where n is the number of row exchanges. (False) (True) Partial and full pivoting is needed to reduce error, where full pivoting reduces ... 2 Since$$\begin{vmatrix} a_{1} & b_{1} & c_{1} \\a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{vmatrix}=a_1b_2c_3+b_1c_2a_3+c_1a_2b_3-a_3b_2c_1-b_3c_2a_1-c_3a_2b_1.$$Each of these terms are either 0 or 1 (depending on the entries chosen to be 1 or 0). So to maximize one may want to choose the entries with positive ... 1 There is a general formula in combinatorics: if you have n objects of which have k_i are of type i, for 1\leq i \leq m, the number of words you can make is \frac{n!}{k_1!k_2!\cdots k_m!}. That's because you first arrange the n objects in one of the n! permutations, then identify all such permutations that are obtained by simply permuting the ... 1 You need to pick 3 positions for the 'a' : 8 \choose 3, then 2 positions for the 'b': 5 \choose 2. Now you are left with the number of words of length 3 on an alphabet of size 2. You have 2^3 words. Finally you have {8 \choose 3} \times {5 \choose 2} \times 2^3=56 \times 10 \times 8=4480 words. 1 The distributing can be done by the following algorithm: First order the objects. This can be done in n! ways. Then place r-1 "dividing sticks" between the objects. There are n-1 gaps where you can place a stick. This will produce a division of the objects to the r boxes such that each box is non-empty. This can be done in {{n-1}\choose{r-1}} ... 1 I will give a symbolic answer, in addition to the well-explained one by Seifert, thanks to which an error has been corrected. Let A=\{1,2,3,5,7,8\}. The sum would be$$\sum_{(a,b,c,d,e,f)\in A^6}\overline{abcdef}=\sum_{(a,b,c,d,e,f)\in A^6}(10000a+10000b+1000c+100d+10e+f)$$and by an rearragement$$=\sum_{(a,b,c,d,e,f)\in A^6}111111a=111111\times(1+2+3+5+...

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