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4 identical dice rolled at the same time what is the number of the possibility to have exactly two 1's ?

somehow i don t find the correct answer between these :

(1):

since it is " at the same time " then the order plays no role here then we have

1122

1123

1124

1125

1126

1133

1134

1135

1136

1144

1145

1146

1155

1156

1166

==> there number of possibility is 15

(2)

since it is " at the same time " then the order plays no role here then we have 11xx and x in [2,6] Then the number of the possibility would be $\binom{2+5-1}{2} = \binom{6}{2} =30 \neq 15 $ !! what was worng !

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    How is $\binom 62=360$? – 2017-01-21
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    because it does not matter which one is the first or the last – 2017-01-21
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    $\binom{6}{2}$ is equal to $15$. http://www.wolframalpha.com/input/?i=6+combination+2 – 2017-01-21
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    I removed that part of my comment because it is specified that the dice are identical. – 2017-01-21
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    $\binom{6}{2} = \frac{6!}{4!} = 30 $ ? – 2017-01-21
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    @MedAyssarBenelhedi That formula is used for permutations. Combinations are given by: $\frac{n!}{(n-r)!r!}=\frac{6!}{4!\cdot 2!}$ – 2017-01-21
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    oh yeah i remember now !! my bad !! i guess i have to take a rest – 2017-01-21
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    yeah you are right ! yeah , just got confused about many things , sorry – 2017-01-21
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    yeah sure , both ways i found the same result – 2017-01-21

1 Answers 1

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Your only mistake is that $\binom{6}{2}$ does not equal to $30$. The formula you used is for permutations: $$^{r+n-1}P_r=\frac{(r+n-1)!}{(r+n-1-r)!}=\frac{(r+n-1)!}{(n-1)!}=\frac{6!}{4!}=30$$ But what you should be looking for is: $$^{r+n-1}C_r=\binom{r+n-1}{r}=\frac{(r+n-1)!}{(r+n-1-r)!r!}=\frac{(r+n-1)!}{(n-1)!r!}=\frac{6!}{4!\cdot 2!}=15$$

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    more accurate was $ \binom{r+n-1}{r}$ – 2017-01-21
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    I guess for other people who may read this , would be better to give them the accurate version – 2017-01-21