How many passwords of 3 distinct digits are possible?
$$10P3=720$$
A 5-member team is formed by two coaches and three players. If in total there are 10 coaches and 15 players, what is the total number of possible teams to be formed.
$$10\times9\times15\times14\times13=245700$$
Do you see any mistake?
On
The first one is correct.
However for the second one, the order each member is picked does not matter. We can pick Paul as the first coach and Rody as the second coach, and that's the same as picking Rody then Paul. For contrast, in the first problem the password 123 is different from 213 since the order makes these two permutations distinct.
So for the second one, the answer is $\binom{10}{2} \times \binom{15}{3} = \frac{10 * 9}{2 * 1} * \frac{15 * 14 * 13}{3 * 2 * 1}$.
This can also be thought of as choosing the two coaches $10 * 9$ ways, then since there are $2$ ways to reorder them, we divide by two since otherwise we double-count. Likewise we have $3 * 2 * 1$ ways to reorder the three players we choose.
Note: $\binom{a}{b}$ is notation for "$a$ choose $b$". It can also be written $ _aC_b$, and is equal to $\frac{a!}{b!(a-b)!}$.
The second question is one of combinations, not of permutations. In particular, when you pick two coaches out of $10$, you pick a combination , not a permutation of the coaches (because once the team is picked, we don't look at the order in which people were picked, but only if they were in the team or not). Therefore, the answer is $^{10}C_2 \times ^{15}C_3$, and NOT $^{10}P_2 \times ^{15}P_3$ which is your answer $245700$.
The first is one of permutations, so your answer is right.