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16

Hint: Once you've picked six different digits, there is only one way to arrange them to satisfy the conditions given.


6

Or to look at it another way. There is only one 9 digit number with this condition 123456789. What about 8 digits?: there are nine answers 12345678, 12345679, 12345689, 12345789, 12346789, 12356789, 12456789, 13456789, 23456789. So all I have done is strike out each digit in turn. $9\times 1$ digit = 9 possibilities. For 7 digits, you need to strike out 2 ...


5

Since the prime factorization of $$210=2\cdot 3\cdot 5\cdot 7$$ What we have to basically do is find out the number of ways we distribute these factors among the four variables $x_1,x_2,x_3$ and $x_4$. We can distribute each factor among the $4$ variables by ${4\choose 1}=4$ ways. Hence, the answer is $$4^4=256$$ Also considering negative numbers, we can ...


3

Note that any two distinct elements $5^{p_1} 13^{q_1} 31^{r_1}$ and $5^{p_2} 13^{q_2} 31^{r_2}$ are incomparable (neither divides the other) if $p_1 + q_1 + r_1 = p_2 + q_2 + r_2$. So a mutually incomparable set is given by $$ S_n = \left\{5^p 13^q 31^r \;\big\vert\; p+q+r=n;\; 0\le p,q,r \le 200\right\} $$ for any $n$. The size of this set is maximal when ...


3

The diagonal line will cross $b$ squares (in the $b$ direction) and $l$ squares (in the ($l$ direction), These ($b$ and $l$ directions) will count separately toward the total number of squares cut except for $\gcd(l,b)$ occasions when the diagonal makes the transition from one row to the next in both directions through a grid point. This counts the start/end ...


3

if $a=1$, for the rest $9$ places, $1$ can be taken anywhere, and also the rest $8$ places must be $0$ or the sum would exceed $2$. Also if $a=2$, then the rest $9$ places must be $0$, or sum would exceed $2$. Hence there are $9+1=10$ such numbers.


3

Hints : The diagonal elements have to be $0$ If the lower left triangle is given, the upper right triangle is determined by $a_{ij}=-a_{ji}$


2

This is the argument by Jack D'Aurizio spelled out. Use the multinomial theorem $$ (x_{1} + \dots + x_{m})^{n} = \sum_{k_1 + \dots + k_{m} = n} \binom{n}{k_1, k_2, \dots, k_{m}} x_{1}^{k_{1}} \cdots x_{m}^{k_{m}} $$ and set all $x_{1} = 1$.


1

The LHS is the number of ways to partion n entities into m sets that can be done by asking individiual element one by one where it wants to go from a list of m subsets that we can form beforehand. Thus ways now would be m for first element, sm for second element and so on. So total ways would be $\prod_1^nm=m^n$.


1

The theoretical minimum is achieved when each square has exactly two red edges and no edge that is in the border is painted red. In this case $64$ edges must be colored. Prove it is possible. Here is a coloring that works:


1

Hint. Put a rook on the board, say on a corner square. Can you make a series of rook moves that traverses every square on the board once and only once, and returns to its starting point? Now what can you say about the set of edges that the rook crosses?


1

For illustrative purposes suppose that there are $3$ types of candy and just the two of us picking. The generating function for my outcomes is simply $1+x_1^2+x_2^2+x_3^2$: the term $1$ corresponds to my picking no candy at all, and for $k=1,2,3$ the term $x_k^2$ corresponds to my picking $2$ candies of type $k$. Those four outcomes are the only possible ...


1

Calling the requested number $f(l,b)$, and putting $d=\gcd(l,b)$, one has $f(l,b)=d f(\frac ld,\frac bd)$. This is because cutting the diagonal into $d$ pieces of equal length, the subdivision points are lattice points (in fact they are precisely the lattice points on the diagonal), and all $d$ pieces are similar. This reduces the problem to the case where ...


1

It is surely sufficient to remove $n$ numbers: we can remove all $n$ even numbers. Then all numbers that are left are odd, so the sum of any two of them is even and in particular not one of the numbers left. To show that we need to remove at least $n$ numbers, consider the largest number $M$ that remains. We distinguish two cases: $M = 2\ell+1$ is odd. We ...


1

Your answer to part 1 is correct. For part 2, you must place the balls so that there is one empty box, one box with two balls, and the remaining balls will have one ball each. There are $n$ ways to pick the empty box, and $n-1$ ways to then pick the box with two balls. We can now fill the $n$ spaces for the balls with the $n$ balls in any order you wish. ...


1

Hint. By your first condition we have $$S=A_1\cup A_2\cup\cdots\cup A_{101}\ .$$ The inclusion/exclusion formula for $101$ sets is $$\eqalign{ N=|S| &=|A_1|+|A_2|+\cdots+|A_{101}|\cr &\qquad {}-|A_1\cap A_2|-\cdots\cr &\qquad {}+|A_1\cap A_2\cap A_3|+\cdots\cr &\qquad {}-|A_1\cap A_2\cap A_3\cap A_4|+\cdots\cr &\qquad ...


1

The stars and bars technique can be used here. There are $\binom{n+k-1}{k-1}$ ways to distribute the balls. Suppose that $1\le m\le k$. There are $\binom{k}m$ ways to choose a subset of the $k$ to receive balls, and there are $\binom{n-1}{m-1}$ ways to distribute the balls to that subset in such a way that everyone receives at least one ball. Thus, the ...


1

number of ways of distributing n identical balls into k persons is given by $n+k-1\choose n$.wiki number of ways of distributing n identical balls into m persons so that each of the m person gets at least one ball and the remaining k-m person gets $0$ ball,is given by ${k \choose m}*{n-1\choose m-1}$. Can you proceed now?


1

Suppose there are $m$ blocks of size $k$ and $n$ of size $3$, then $3n=\binom{v}{2}-m\binom{k}{2}$, notice $\binom{v}{2}$ is congruent to $1\bmod 3$. Unless $k$ is congruent to $2\bmod 3$ we shall have $\binom{k}{2}$ is a multiple of three. This would mean $\binom{v}{2}-m\binom{k}{2}$ would not be a multiple of $3$, and $n$ would then not be an integer.



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