Related to the partition of $n$ where $p(n,2)$ denotes the number of partitions $\geq 2$ I was trying the following question from the Number theory book of Zuckerman. 
If $p(n,2)$ denotes the number of partition of n with parts $\geq2$, prove that $p(n,2)>p(n-1,2)$ for all $n\geq 8$, that $p(n)=p(n-1)+p(n,2)$ for all $n \geq 1$, and that $p(n+1)+p(n-1)>2p(n)$ for all $n \geq 7$.
Need some clue to solve it.
 A: The technique I will be using may be summarized thus: To show $a<b$ I prove that there is an injection from a set of $a$ elements to a set containing $b$ elements which is not a surjection.


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*Every partition of $n-1$, say $n-1=\alpha_1+\cdots \alpha_k$, where $ \alpha_k\ge\alpha_{k-1}\ge\cdots\ge 2$ induces a partition of $n$ with each part $\ge 2$ in a natural fashion: $n=\alpha_1+\cdots+(\alpha_k+1)$. Moreover such distinct partitions of $n-1$ yield different such partitions of $n$. This gives us an injection from the set of all partitions counted in $p(n-1,2)$ to the set of all partitions counted in $p(n,2)$. Furthermore, this map is not a surjection for the following reason: If $n=2t$ is even then nothing covered in $p(n-1,2)$ induces $n=\underbrace{2+2+\cdots 2}_{\text{t times}}$. If $n$ is odd then $n=\frac{n-3}{2}+\frac{n-3}{2}+3$ is not induced by anything, because its supposed preimage, namely $n-1=\frac{n-3}{2}+\frac{n-5}{2}+3$, instead induces $n=\frac{n-1}{2}+\frac{n-5}{2}+3$. This is under the assumption that $\frac{n-5}{2}\ge 2$, i.e. $n\ge 9$. Combining this with the even case we see that for strict inequality $n\ge 8$. The result follows.

*Any partition of $n$ either has a $1$ in it or not. If it does, then by removing the $1$ we get a partition counted in $p(n-1)$. If it does not, then we have a partition counted in $p(n,2)$. The converse holds trivially and so we have a bijection from the set of all partitions of $n$ to the disjoint union of all partitions of $n-1$ and partitions of $n$ with each part at least $2$.

*Part 2 gives us a formula for $p(n,2)$, viz $p(n,2)=p(n)-p(n-1)$. Substituting in part 1 and rearranging gives us $p(n)+p(n-2)<2p(n-1)$ for $n\ge 8$. Let $N+1=n$. Then $p(N+1)+p(N-1)<2p(N)$ for $N\ge 7$ which establishes the result.
