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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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