How many 5-digit numbers are there such that the ten-thousand's digit is equal to the sum of the other 4 digits?

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Details and Assumptions:
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- $70340$ is one such number
- The ten-thousand digit can not be $0$

The answer is 714.

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## Defining Notations

Let the 5-digit number be of the form $abcde$ . Since $1 \leq a \leq 9$ , we can make cases when $a = 1, 2, \ldots 9$

## Making Cases

We can make cases when $(b + c + d + e = 1, 2, \ldots 9)$ .

When $b + c + d + e = 1$ , using the Stars and bars theorem the number of possibilites is $\binom{1 + 4 - 1}{4 - 1} = \binom{4}{3}$

When $b + c + d + e = 2$ , using the same techinque the number of possibilites is $\binom{2 + 4 - 1}{4 - 1} = \binom{5}{3}$

When $b + c + d + e = 3$ , using the same techinque the number of possibilites is $\binom{3 + 4 - 1}{4 - 1} = \binom{6}{3}$

We can continue to list down all the possibilities till $b + c + d + e = 9$ . This equals $\binom{12}{3}$

## Adding all Combinations

$\begin{aligned}\text{Total such Numbers} &= \binom{4}{3} + \binom{5}{3} + \binom{6}{3} + \ldots + \binom{11}{3} + \binom{12}{3} \\ &= \binom{13}{4} - \binom{3}{3} \\ &= 715 - 1 \\ &= \boxed{714} \end{aligned}$