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 ?
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.
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.