$φ(n)=(p-1)(q-1)(r-1)$ Let $p, q, r$ be three distinct primes. Show that $φ(n)=(p-1)(q-1)(r-1)$
So far I have:
There are $qr-1$ multiples of $p$ in $1,...,pqr$
There are $pr-1$ multiples of $q$ in $1,...,pqr$
There are $pq-1$ multiples of $r$ in $1,...,pqr$
We counted $pqr$ 3 times
$p$ & $q$ share $r-1$ multiples
$q$ & $r$ share $p-1$ multiples
$p$ & $r$ share $q-1$ multiples
Therefore, $φ(n)=pqr-(qr-1)-(pr-1)-(pq-1)+(r-1)+(p-1)+(q-1)-2$
Which does not simply down to  $φ(n)=(p-1)(q-1)(r-1)$
What am I missing? Thanks for any help! 
 A: What you are missing is that it does simplify down to $(p-1)(q-1)(r-1)$:
\begin{align*}
(p-1)(q-1)(r-1) &= pqr -qr-pr-pq+p+q+r-1\\
&=pqr-(qr-1)-(pr-1)-(pq-1) -3  \\
&\qquad +(p-1)+(q-1)+(r-1)+3 -1\\
&=pqr-(qr-1)-(pr-1)-(pq-1) \\
&\qquad+ (p-1)+(q-1)+(r-1) -1.
\end{align*}
A: Alternate approach: For any prime $p$, we have $\phi(p)=p-1$. 
Then use the fact that $\phi$ is multiplicative.  That is, $\operatorname {gcd}(m,n)=1\implies \phi(mn)=\phi(m)\phi(n)$. (See the Chinese remainder theorem for a proof of this property.)
A: Your error seems to be that you count up to but not including $pqr$.
The multiples of $p$ are $p,2p, 3p,.....,(qr-2)p,(qr-1)p,$  AND $pqr$.  So there are $qr$ multiples; not $qr -1$.
If we redo those numbers, inclusion-exclusion gives us $pqr - pq-qr-pr +p+q+r -1$ which is $(p-1)(q-1)(r-1)$.
If we omit $pqr$ and say there are $qr-1$ multiples of $p$ in $1,....,pqr-1$ then we would, by inclusion-exclusion have:
$(pqr-1) - (pq-1) -(pr-1)-(qr-1) + (p-1)+(q-1)+(r-1) - 0$ which is the same result.
