<s>[INST] Number of ways of getting same number when four people throw a die once Four people are rolling a die once. How many ways


*

*None of them get same number

*Exactly two of them get same number

*Two of them get the same number

*Three of them get same number

*All of them get same number


Solution: 


*

*None of them get same number$= 6 \cdot 5 \cdot 4 \cdot 3= 360$ ways

*Exactly two of them get same numbers$= 4C2 \cdot 6 \cdot 5 \cdot 4= 720$ ways

*Two of them same numbers$= 4C2 \cdot 6 \cdot 5 \cdot 5= 900$ ways

*All of them get same number$= 4C4 \cdot 6 \cdot 1 \cdot 1= 6$ ways
Is my approach correct? Can anyone clarify in detail?
 [/INST] Your answers to the first two questions are correct, as is your answer to the last question.
If we interpret the third question to mean at least two of them get the same number, then we can compute the answer by subtracting the number of cases in which none of them get the same number from the total number of outcomes.  Since there are six possible outcomes for each of the four times the die is thrown, the total number of possible outcomes is $6^4$.  You found that the number of outcomes in which no two people get the same number is $6 \cdot 5 \cdot 4 \cdot 3$.  Hence, the number of outcomes in which at least two people get the same number is 
$$6^4 - 6 \cdot 5 \cdot 4 \cdot 3 = 1296 - 360 = 936$$
We can verify this by considering the cases in which at least two people obtain the same number.  Since the partitions of $4$ are
\begin{align*}
4 & = 4\\
  & = 3 + 1\\
  & = 2 + 2\\
  & = 2 + 1 + 1\\
  & = 1 + 1 + 1 + 1
\end{align*}
the set of possible cases is 


*

*no two people get the same number

*exactly two people get the same number

*two distinct numbers occur, with two people apiece getting each outcome

*exactly three people get the same number

*all four people get the same number


You correctly calculated that the number of cases in which exactly two people get the same number is 
$$\binom{6}{2} \cdot 6 \cdot 5 \cdot 4$$
The number of cases in which two distinct numbers occur, with two people apiece getting each outcome is 
$$\binom{6}{2}\binom{4}{2} = 90$$
since there are $\binom{6}{2}$ ways of selecting two numbers and $\binom{4}{2}$ ways for exactly two of the four people to get the larger of those numbers.
The number of cases in which exactly three people get the same number is 
$$\binom{4}{3} \cdot 6 \cdot 5 = 120$$
since there are $\binom{4}{3}$ ways for exactly three of the people to get the same number, six choices for the repeated number, and five choices for the remaining number.  
Since there are six possible outcomes in which all four people get the same number, the number of outcomes in which at least two people get the same number is 
$$\binom{4}{2} \cdot 6 \cdot 5 \cdot 4 + \binom{6}{2}\binom{4}{2} + \binom{4}{3} \cdot 6 \cdot 5 + \binom{4}{4} \cdot 6 = 720 + 90 + 120 + 6 = 936$$
as we found above.
If we interpret the fourth question to mean at least three people get the same number, we can find the answer by adding the cases in which exactly three people get the same number and exactly four people get the same number.
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