How many solutions does the rebus have? How many 8 digit (its digits from left to right are labeled a through g) satisfy the following constraints:


*

*$a_1 < a_2 < a_3 < a_4$

*$a_4 > a_5$
 A: First set $f$:
$$f=9 \Longrightarrow 4C_5^9=6C_6^9$$
$$f=8 \Longrightarrow 3C_5^8=6C_6^8$$
$$f=7 \Longrightarrow 2C_5^7=6C_6^7$$
$$f=6 \Longrightarrow 1C_5^6=6C_6^6$$
Since the left sides of equations produce when you first choose $a,...,e$, then choose $g$ of remaining digits. And the right sides produse when you choose $a,...,e$ and $g$, then choose $g$ from them. At end we can compute the number of solutions is $720$.
A: Hint:
First choose $f\in\{6,7,8,9\}$
Then choosing $abcde$ boils down to selecting $5$ distinct digits from $\{1,\dots,f-1\}$.
And choosing $g$ boils down to selecting a digit from $\{1,\dots,f-1\}$.
A: I am assuming that zero is not a valid first digit, so all of $a$ through $g$ are digits in the range $1$ through $9$ (inclusive).
For the moment, let's disregard the constraint $f > g$. Then there are $\binom{9}{6}$ ways to chose $a,b,c,d,e,f$ and $9$ ways to choose $g$, for a total of $N_{tot}=\binom{9}{6} \cdot 9 = 756$ possible 7 digit numbers.
Now either $f<g$, $f=g$, or $f > g$.  


*

*If $f<g$, then there are $N_{f<g} = \binom{9}{7}=36$ possible 7 digit numbers.  

*If $f=g$, then there are $N_{f=g} = \binom{9}{6}=84$ possible 7 digit numbers.


But $N_{tot} = N_{f<g} + N_{f=g} + N_{f>g}$, so the number of 7 digit numbers with $f>g$ is
$$N_{f>g} = N_{tot} - N_{f<g} - N_{f=g} = \boxed{636}$$
