How many $7$ digit numbers in the form $a_7a_6...a_1$ such $\forall i \in (1,6)$ $a_{i+1}\geq a_i$, $0$ leading not allowed.
It is easy if the condition was $\forall i \in (1,6)$ $a_{i+1} > a_i$, I have recognized that I have $10$ ways to put $a_1$ the next $a_2$ has $C_{10-a_1}^1$ ways and so on. can you help me please.
Thank you.
For such a number, the $a_i+i$ form a srtictly increasing sequence of numbers $\in\{1,\ldots, 16\}$, and this can be identified with a size-$7$ subset of this size-$16$ set. Hence there are $16\choose 7$ such numbers, starting from $0000000$ corresponding to $\{1,2,3,4,5,6,7\}$ and running up to $9999999$ corresponding to $\{10,11,12,13,14,15,16\}$. Only one case of leading zero (namely $0000000$) needs to be removed, so the final answer is $${16\choose 7}-1.$$
Although it has been clarified that the "alternative" condition was intended, I shall leave this answer to the wrong problem.
If $i=7$ then there is no $a_{i+1}$ so it seems to me that the condition $\forall i \in (1,7) \text{ } a_{i+1} \ge a_i$ is badly written. I suggest this should be $\forall i \in (1,6) \text{ } a_{i+1} \ge a_i$.
The way I interpret the condition is that $a_7 \ge a_6$ and $a_2 \ge a_1$. This means that $a_5, a_4, a_3$ can be any digits.
(Alternatively it might be intended that the digits do not increase from left to right : $\forall i \in (1, 2, 3, 4, 5, 6)=(1..6) \text{ } a_{i+1} \ge a_i$.)
So there are 3 independent problems :
(a) in how many ways $P_a$ can you write a 2-digit number $a_7a_6$ such that $a_7 \ge a_6$ with $a_7 \ne 0$?
(b) in how many ways $P_b$ can you write a 3-digit number $a_5a_4a_3$ with no restrictions (so $a_5$ can be $0$)?
(c) in how many ways $P_c$ can you write a 2-digit number $a_2a_1$ with $a_2 \ge a_1$ (again $a_2$ can be $0$)?
Finally combine all 3 results : $P=P_a \times P_b \times P_c$.
