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6

It is impossible. Consider the colouring of the board below. Suppose it is possible. Then each $2 \times 2$ piece covers exactly one black square and each $1 \times 4$ piece covers either zero or two black squares. But there are $64$ black squares and $64 - 33$ is odd, contradiction.


5

One way is to consider putting some/all/none of the chocolates on the left and the rest on the right. Each chocolate can go either way, so there are $2^n$ possibilities. But $2$ of these possibilities leave one side empty. Ignoring these, there are $2^n-2$ possibilities. Each pattern has a distinct reflection, swapping all those on the left with all ...


5

use the hockey stick identity. What we want is $\sum_{i=3}^{n}\binom{n}{3}6$ It is a known fact $\sum_{i=k}^n\binom{i}{k}=\binom{n+1}{k+1}$ Hence $\sum_{i=3}^{n}i(i-1)(i-2)=6\binom{n+1}{4}$


3

We want to group $12$ people into $6$ single-sex groups of $2$, so I began by calculating the number of ways we can make $6$ pairs. By the multinomial function: $12!/(2!)^6=7484400$ ways to make $6$ pairs. No, order doesn't matter within the groups of $2$ nor of the $6$ groups. $$\frac{12!}{2!^6 6!} = 10395$$ Then, we want to figure out the ...


2

Consider the simulataneous inequalties $$a + b + c + d > 0,$$ $$a + b + c < 0,$$ $$b + c + d < 0.$$ What can we say about the sign of $a$? What about the sign of $d$? If you have a sequence of six numbers, it has three subsequences each consisting of four consecutive numbers. Consider what each of them tells you about numbers in the original ...


2

$0>(a+b+c)+(b+c+d)+(c+d+e)+(d+e+f)=a+2 b +3 c +3 d +2 e+f.$ $0<(a+b+c+d)+(b+c+d+e)+(c+d+e+f)=a+2 b+3 c+3 d+2 e +f.$


2

The conjecture is true if no non-empty proper subset of $\{a_1,\ldots,a_n\}$ sums to $0$. Hereafter I will make that assumption. Let $\langle b_0,\ldots,b_{n-1}\rangle$ be any permutation of the numbers $a_1,\ldots,a_n$. Extend it to an infinite periodic sequence $\langle b_k:k\in\Bbb N\rangle$ with period $n$ by setting $b_{\ell n+k}=b_k$ for ...


2

Using the Lemma, it can be seen that $\dbinom{m}{r}$ is a positive integer, since, $\dbinom{m}{n} = \dfrac{(m)!}{n!(m-n)!}$ Also, since $\displaystyle \dbinom{m}{r} = \dfrac{m}{r} \dbinom{m-1}{r-1}$, $$\implies \dfrac{1}{r} \dbinom{rm}{m} = \dbinom{rm-1}{m-1} \tag{1}$$ Thus, by $(1)$, $\displaystyle \dfrac{1}{r}\dbinom{rm}{m} \in ...


2

Need to follow hint your professor said, calculate a few terms, and factor out n: $$k\binom nk = k \frac{n!}{k!(n-k)!} = n\frac{(n-1)!}{(k-1)!(n-k)!} = n\frac{(n-1)!}{(k-1)!((n-1) - (k - 1))!} = n\binom {n-1}{k-1}$$ Follow the rest hints to finish the proof. Now we use the hint $$\sum_{j=0}^m \binom{m}{j} = 2^m$$ $$\sum_{k=1}^n k \frac{n!}{k!(n-k)!} = ...


1

I will give an inductive proof using the fact that $ \sum_{k=0}^n \tbinom n k = 2^n$ as well as Pascal's identity. The base case is straightforward. Suppose it holds for some $n$. Then $$\sum_{k=1}^{n+1} k\binom {n+1} k = n+1 + \sum_{k=1}^{n} k\binom {n} {k-1} + \sum_{k=1}^{n} k\binom {n}{k} = n + 1 + \sum_{k=1}^{n} (k-1)\binom {n} {k-1} + ...


1

Its just simply cases 1) no orange is selected ${5\choose 4}=5$ .2)when one orange is selected is $1.{5\choose 3}=10$ and 3)both oranges are selected so $1.1.{5\choose 2}$ so total ways are $5+10+10=25$


1

Add up the following: The number of combinations containing $0$ oranges, which is $\binom54=5$ The number of combinations containing $1$ oranges, which is $\binom53=10$ The number of combinations containing $2$ oranges, which is $\binom52=10$



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