Throwing a die 4 times and calculating the number of outcomes

Question

Okay, so this is my problem: How many outcomes where exactly ONE number occurs exactly TWO times ex (2, 1, 2, 5) or (3, 6, 6, 1) but NOT (1, 2, 1, 2) or (3, 1, 3, 3)

My solution: So the first throw has 6 possibilities, then this must be followed by one throw with only ONE possibility, then followed by a throw with 5 possibilities, followed by a throw with 4 possibilities. Then this gives me: 6 * 1 * 5 * 4 = 120 outcomes

This solution gave me 0 points, so my question is where do I go wrong?

Answer 1

Forget the dice, suppose you just had to pick four numbers from 1 to 6 , and you can repeat numbers. For your case, we would need to firstly pick the number that is going to be appearing twice - there are 6 ways to do it

Now, we need to pick two distinct numbers from the remaining 5 - there's ${5 \choose 2}$ ways of doing that

Now for order - we have a sequence of the type $A A B C$ - hence the total number of permutations of this list is $\frac{4!}{2!}$

Put it all together

$$N = 6 \times {5 \choose 2}\times \frac{4!}{2!}$$

Probability = $\frac{N}{6^4}$

Answer 2

Although a correct solution has been given assuming it isn't a die, and the problem is simple enough to be solved in an elementary way, it would be good for you to know how to solve a bigger die problem in an almost mechanical way as the product of two multinomial coefficients.

Suppose the problem is to find in how many ways a die rolled $11$ times yields two triples, one double and three singles, ie of the pattern $3-3-2-1-1-1$

Of the $6$ "slots", we need to assign two for the $3s$, one for the $2s$ and three for the $1s$ which is given by the multinomial coefficient $\dbinom{6}{2,1,3}$

And we need to permute the result to account for multiple hits in a slot which is given by the multinomial coefficient $\dbinom{11}{3,3,2,1,1,1}$

so answer = $\dbinom{6}{2,1,3}\dbinom{11}{3,3,2,1,1,1}$

You may not yet have learnt of the various interpretations of the multinomial coefficient, but arithmetically it is essentially permutations with repeated elements, which you must have learnt, so you can simply use $\;\;\dfrac{6}{2!1!3!}\cdot\dfrac{11!}{3!3!2!1!1!1!}$

I hope this is of help to you as you advance into more complex problems of this type.