Mistakes Give Rise to Problems- 18

See that because $A,B,C,D,E,F,G,H,I,J$ are distinct integers,

the minimum value of $A+B+C+D+E$ will be for

$(A,B,C,D,E)$ = some permutation of $(1,2,3,4,5)$ because these are the minimum distinct positive integers. Hence their sum can be minimum $15$ , and it will be $15$ in only this case.

Because it is given to be $15$ , we can conclude that $(A,B,C,D,E)$ is some permutation of $(1,2,3,4,5)$ . (giving $5!=120$ ordered 5-tuples )

Now because $F,G,H,I,J$ are distinct (other than A,B,C,D,E), they can have minimum sum by $(F,G,H,I,J)$ = some permutation of $(6,7,8,9,10)$ . And thus the minimum sum they can give will be $6+7+8+9+10=40$ .

This gives us that $(F,G,H,I,J)$ is some permutation of $(6,7,8,9,10)$ . (giving $5!=120$ ordered 5-tuples )

Thus the total number of $10-tuples$ will be $5!\times 5!$ (by rule of product) giving the final answer to be $\boxed{14400}$