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3 votes

Let $n \geq 2$ be any integer. Which of the following statements is not necessarily true?

- $\begin{pmatrix} n \\ i \end{pmatrix} = \begin{pmatrix} n-1 \\ i \end{pmatrix} + \begin{pmatrix} n-1 \\ i-1 \end{pmatrix}, \text{ where } 1 \leq i \leq n-1$
- $n!$ divides the product of any $n$ consecutive integers
- $\Sigma_{i=0}^n \begin{pmatrix} n \\ i \end{pmatrix} = 2^n$
- $n$ divides $\begin{pmatrix} n \\ i \end{pmatrix}$, for all $ i \in \{1, 2, \dots , n-1\}$
- If $n$ is an odd prime, then $n$ divides $2^{n-1} -1$

1 vote

A,B,C are true due to usual reasons.

Now,

Let a = 2 which is not divisible by n, then using Fermat's little theorem

$2^{n-1}\equiv 1(mod n)$

$\Rightarrow$ $2^{n-1} -1\equiv 0(mod n)$

Therefore **E** is true.

**D.** But it also seems to be correct within given range.

For n= 1 : n divides $\frac{n(n-1)}{2}$ and so for n-1.

For i = k, k between 1 & n-1. n divides $\frac{n(n-1)(n-2)...(n-k+1)}{k!}$ , Since there is n at numerator always. Therefore True.

0 votes

$D$

If $a$ divides $b$$\Rightarrow a\,\,|\,\,b\Rightarrow \,b=a*c$ for some integer $c$

$n$ divides $\begin{pmatrix} n \\ i \end{pmatrix}$, for all $ i \in \{1, 2, \dots , n-1\}$

take $n$=even,

say $n=6,i=2$,

$\binom{6}{2}=15$

$15\neq c*6$ for any $c$.

Hence $D$ false.

Even in the answer key,it is the answer

If $a$ divides $b$$\Rightarrow a\,\,|\,\,b\Rightarrow \,b=a*c$ for some integer $c$

$n$ divides $\begin{pmatrix} n \\ i \end{pmatrix}$, for all $ i \in \{1, 2, \dots , n-1\}$

take $n$=even,

say $n=6,i=2$,

$\binom{6}{2}=15$

$15\neq c*6$ for any $c$.

Hence $D$ false.

Even in the answer key,it is the answer

0 votes

a) It's Pascal's Identity Theorem

https://en.wikipedia.org/wiki/Pascal%27s_rule

b) It's quite tricky. We know $nCr$ is always an integer.

$nCr$= $\frac{n!}{r!*(n-r)!}$ = $\frac {r! * (r+1)(r+2)...n}{(n-r)!*r!}$

NOw observe this carefully $(r+1)(r+2)...(n)$ are product of any consecutive $(n-r)$integers and this would be divisible by $(n-r)!$ because $nCr$ is always an integer.

c) Quite basic. $nC0+nC1+nC2+....nCn$=$2^n$

d) If $n$ and $r$ are relatively prime then certainly $nCr$ is divisible by $n$.When they are not co prime we can't claim anything.