# Tagged Questions

188 views

### Wheel of Fortune Problem

The Summation formula is $$\sum_{i=1}^ni =\frac{n(n+1)}2$$ How is it that we know the integers $1,2,...36$ appear exactly $3$ times. And why do we multiply the sum by $3$ in the last part of the ...
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### Winning or Non-losing strategy for A or B

Find a winning or a non-losing strategy for the following game: Consider $25$ sticks arranged in a $5$ x $5$ square. Players alternately take any number of sticks from a single row or column. At ...
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### Combinatorics Proof

I am having trouble with a combinatorics proof. I need to prove that if $r$ <= $n$ then the number of $r$ - subsets of {1,...,n} is $n!$/$(n-r)!$*$r!$ I really struggle with writing proofs and ...
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### Show that $C(n,k) = C(n-1,k) + C(n-1,k-1)$ [duplicate]

I'm studying for my final for Statistics, and I want to understand literally every problem in my textbook (at least in the first 7 chapters). One of the problems asks to show that ${n}\choose {k}$ ...
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### How do I begin proving this binomial coefficient identity?

This is a homework question. I'm asked to prove the identity: $${n\choose 0} - {n\choose 1} + {n\choose 2} - {n\choose 3} + \dots = 0$$ (The sum ends with ${n\choose n} = 1$, with the sign of the ...
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### Combinatorics proof of “sum of (k choose m) with k from m up to n is equal to n+1 choose m+1”

I've already proved this statement algrebraically. I'm asked to prove it with combinatorics. So far I came up with, LHS= # ways to choose m apples from a total of m,m+1,...,n RHS= # ways to choose ...
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### Prove: Number of Derangement is odd if and only if number of items is even .

let $D_n$ be a number of Derangement of n items . prove that $D_n$ is odd if and only if n is even . i was trying to use induction on the $!n=(n-1)(!(n-1)+!(n-2))$ recurrence relation but i cant ...
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### Proof of a binomial identity $\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$

Prove that $$\sum_{k=0}^n {n \choose k}^{\!2} = {2n \choose n}.$$ The exercise provides the following hint: $\,\,\displaystyle{n \choose k}={n\choose n-k}$. Any help?
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### Sum of $k {n \choose k}$ is $n2^{n-1}$

Proof that $\suṃ̣_{k=1}^{n}k {n \choose k}$ for $n \in \mathbb N$ is equal to $n2^{n-1}$. As a hint I got that $k {n \choose k} = n {n-1\choose k-1}$. I tried solving this by induction but, in the ...
Problem: Show that these three statements are equivalent, where $a, b \in R:$ (i) $a < b$, (ii) the average of $a, b,$ is greater than $a,$ and (iii) the average of $a$ and $b$ is less than $b$. ...