SOLUTION: A restaurant has 3 kinds of toys and is giving one toy for each purchase of a kiddie meal. Find the probability of getting three DIFFERENT toys in five purchases.

The first answer I gave you was incorrect.  I regret this.
Please ignore it and accept my apology.  Here is the correct 
solution: 

We will first find the probability of the complement event,
where there are only 1 or 2 types of toys.  Then we will
subtract from 1 to find the desired probability. 

Let the 5 purchases be purchases 1,2,3,4, and 5, say in
order of the earliest purchase to the latest purschase.

Case 1. 3 purchases getting one kind of toy and 2 purchases
        getting another kind of toy.
Case 2. 4 purchases getting one kind of toy and 1 purchase
        getting another kind of toy.
Case 3. 5 purchases getting the same kind of toy all 5 times.
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Now we calculate the number in each of the above 3 cases.

Case 1. 3 purchases getting one kind of toy and 2 purchases
        getting a second kind of toy.

Choose the 3 purchases for getting one kind of toy.
That's 5 purchases choose 3 or 5C3 ways.
Choose the remaining 2 purchases for getting the  
second kind of toy.
That's 2 purchases choose 2 or 2C2 ways.
Choose the type toy for the three of one kind 3C1 ways
Choose the type toy for the two of another kind in 2C1 ways   

For case 1, 5C3×2C2×3C1×2C1 = 10×1×3×2 = 60 ways

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Case 2. 4 purchases getting one kind of toy and 1 purchase
        getting another kind of toy

Choose the 4 purchases for getting one kind of toy.
That's 5 purchases choose 4 or 5C4 ways.
Choose the remaining 1 purchase for getting the  
other kind of toy.
That's 1 purchase choose 1 or 1C1 ways.
Choose the kind of toy for the four purchases getting the 
same kind of toy.
That's 3 toy types choose 1 or 3C1
Choose the toy type for the two purchases getting the 
other kind of toy.
That's 2 toy types choose 1 or 2C1 

For case 2, 5C4×1C1×3C1×2C1 = 5×1×3×2 = 30 ways 

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Case 3. 5 purchases getting the same kind of one toy 
in each purchase.       

That's 3 toy types choose 1, or 3C1

For case 3, 3C1 = 3

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Total for all three cases: 60+30+3 = 93

So the numerator for the probability of the complement
event is 93

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Now we calculate the denominator for the probability.
That's all the ways the toy types can come up.

Choose the toy type to get in the 1st purchase in 3 ways.
Choose the toy type to get in the 2nd purchase also in 3 ways.
Choose the toy type to get in the 3rd purchase also in 3 ways.
Choose the toy type to get in the 4th purchase also in 3 ways.
Choose the toy type to get in the 5th purchase also in 3 ways.

That's 3×3×3×3×3 = 35 = 243 ways.

So the probability of the complement event is 93/243 = 31/81.

Therefore the probability of the desired event is 

1 - 31/81 = 81/81 - 31/81 = 50/81 

THAT's the correct answer.  Sorry I got it wrong the first time.

Edwin

This is me, Edwin, under my alias "AnlytcPhil".  I have 
this alias so that if I need to correct a problem that I 
have submitted I can do so and the student will be notified.
The correct solution is above.  Again I regret submitting
an incorrect solution.  

Edwin McCravy