How many possible outcomes are there when 5 similar dices are rolled? In this question If I consider first total no of outcomes as $6^5$ , then I divided it be ${6\choose5} *5!$ since there are 5 similar dices so the outcome (1 2  3 4 5 ) will be similar to (5 4 3 2 1) .
Now what's wrong with this approach ?
 A: The answer is $6 + 5 - 1 \choose 5$. The reason is, to exclude re-counting equivalent outcomes, we look at the number of dices that land on $1, 2, 3, 4, 5, 6$. Now, there are $6$ possibilities. There are $5$ dices. Initially, you might think "Aha! The answer is just $6 \choose 5$." But it's a little more complicated than that. Because two dices could land on the same number. So we use the well known, tried and tested, stars and bars method see stars and bars to count allowing for these conincidences, and the answer is as above.
A: Answer expanded
Choosing various combos of numbers, and permuting them in the sequence, we get
$5$ of a kind: $\binom61*\frac{5!}{5!} = 6$
$4-1$ of a kind: $\binom61\binom51*\frac{5!}{4!}=150$
$3-2$ of a kind: $\binom61\binom51*\frac{5!}{3!2!} = 300$
$3-1-1$ of a kind: $\binom61\binom52*\frac{5!}{3!}=1200$
$2-2-1$ of a kind: $\binom62\binom61*\frac{5!}{2!2!} = 1800$
$2-1-1-1$ of a kind: $\binom61\binom53*\frac5{2!} = 3600$
$1-1-1-1-1$ of a kind: $\binom65$*5! = 720
It adds up to 7776 = 6^5, as it must
Conclusion
There are $7$ types of outcomes adding to $252$ (which could have been obtained more simply using "stars and bars",) but the types are not equiprobable, and their distribution over an equiprobable sample space has been worked out, as the question is not very clear as to what exactly is meant here by an outcome
A: You are assuming 5 distinct values. There are more than a few cases to consider here assuming 6 sided dice where things can get tricky:


*

*5 distinct values. In this case there are 6 possible combinations as it is just a matter of which value isn't showing up though there are 120 ways to roll the 5 dice to get that outcome.

*A pair with 3 distinct other values, e.g. 1 1 2 3 4.

*3 of a kind with 2 distinct other values, e.g. 1 1 1 2 3.

*A pair of pairs with one other value, e.g. 1 1 2 2 3.

*A pair with 3 of a kind, e.g. 1 1 1 2 2.

*4 of a kind with one other value, e.g. 1 1 1 1 2.

*All 5 dice show the same value which would be known as a Yahtzee! in that game.
These are all different cases that you'd have to add together.
A: My explanation is that the probability of 1 set of the same value is 1/(number of possible combinations) simply. We are talking about 5 dice with six possible values so the calculation would be 1/(6^5).
In more detail, there are 6 combinations for 1 dice and there are 5 dice so you times the 6 combos by itself 5 times to get to the total combinations possible in the set.
6^5=7776 so the answer is 1/7776.
To calculate the probability from a certain number of throws  add the probabilities of separate throws. Let’s say there is one combination (1, 1, 1, 1, 1). We already know that the probability of getting this is 1/7776, but what about in two throws?
In two throws getting 5 of a kind (for example: 1, 1, 1, 1, 1) you can think of two sets of combinations which give the value of one single combination (for example: 1,1 and 1,1,1). The probability of each part added will give the total and you will find that for this particular example it is 1/250 because it is 1/((6^2)+(6^3)).
However, for the second possible case (1, and 1,1,1,1) the probability is lower because increasing powers don’t follow a sequence and four 1s in one throw has a much lower probability than 1,1,1. This case has a probability of 1/1302.
