Let $\mathbb F$ be a field of $p$ elements.Let $V$ be an $n$-dimensional vector space over $\mathbb F$. Prove the following $:$
$(a)$ $V$ has $p^n$ elements.
$(b)$ $V$ has $p^n - 1$ linearly independent singleton sets.
$(c)$ The number of linearly independent subsets of $V$ consisting of $m$ elements,$(1 \leq m \leq n)$ is
$$ \frac {1} {m!} \prod_{k=0}^{m-1} (p^n - p^k).$$
Hint $:$ Induction.
$(a)$ follows from the fact that $V \simeq \mathbb F^n$ and $\mathbb F$ has $p$-many elements. Also $(b)$ follows immediately from $(a)$. Now to show $(c)$ we use induction on the number of elements in a linearly independent subset of $V$. Clearly our result holds for $n=1$ by part $(b)$. Let our result be true for any linearly independent subset of $m-1$ elements i.e. number of linearly independent subsets of $V$ consisting of $m-1$ elements is
$$ \frac {1} {(m-1)!} \prod_{k=0}^{m-2} (p^n - p^k).$$
We have to prove that our result is true for $m$. Now we can make a linearly independent subset consisting of $m$ elements from a linearly independent subset consisting of $m-1$ elements by adjoining an element of $V$ which is not in the span of these $m-1$ elements. Now the number of elements in the span of $m-1$ elements is $p^{m-1}$. So the adjoining elements should then be chosen from the remaining $p^n - p^{m-1}$ elements and this is true for each of these $\frac {1} {(m-1)!} \prod_{k=0}^{m-2} (p^n - p^k)$ possibilities. So the total number of linearly independent subsets is thus $$ \frac {1} {(m-1)!} \prod_{k=0}^{m-1} (p^n - p^k).$$ which is not the required answer we are looking for. I am quite unsure about where I have done mistake?
Please help me in this regard. Then it will be very helpful for me.
Thank you in advance.
In your counting you state that every LI set of size $m$ is obtained by adjoining an element of $V$ to an LI set of size $m - 1$, which is true. However there isn't a unique way to do this.
If $\{v_1, \ldots, v_m\}$ is an LI set then there is $\binom{m}{1} = m$ different ways to partition $\{v_1, \ldots, v_m\}$ into a set of size $m - 1$ and a single element. You can add $v_1$ to $\{v_2, \ldots, v_m\}$, or $v_2$ to $\{v_1, v_3, \ldots, v_m\}$, and so on. Hence your method of counting constructs the set $\{v_1, \ldots, v_m\}$ $m$ different times.
To compensate for this we divide by $m$ at the end, which will give you the right answer.