In a clairvoyance test, a stack of 25 cards is used, with each of the 5 types of cards occurring 5 times.
You take 3 cards at once. In how many ways can you not take cards of type 1 and type 2?
Solution:
There is no difference between cards of the same type. The possible cards that you can take are:
type 3 - type 3 - type 3
type 4 - type 4 - type 4
type 5 - type 5 - type 5
type 3 - type 3 - type 4
type 3 - type 3 - type 5
type 4 - type 4 - type 3
type 4 - type 4 - type 5
type 5 - type 5 - type 3
type 5 - type 5 - type 4
type 3 - type 4 - type 5
I think there are 10 possibilities. However, my book provides ncr (15,3) as a solution. What is correct?
The book is correct.
To see where you went wrong, consider the following outcomes . . .
The number of ways to get three $3$'s is ${\large{\binom{5}{3}}}=10$.
The number of ways to get two $3$'s and one $4$ is ${\large{\binom{5}{2}}}{\large{\binom{5}{1}}}=50$.
The number of ways to get one $3$, one $4$, and one $5$ is ${\large{\binom{5}{1}}}^3=125$.
So these three outcomes are not equally numerous as $3$-card hands, but in your analysis, you gave them the same weight.
The key is that the $25$ cards cards are physically distinct, so two distinct cards of the same rank should not be regarded as the same card.
Thus, since there are $15$ cards with ranks $3$, $4$, or $5$, the number of $3$-card hands from those $15$ cards is ${\large{\binom{15}{3}}}$.