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A dice is tossed seven times. What is the probability that all six faces appear at least once?
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The dice is tossed $7$ times . It has 6 faces so possible sample space is $\large 6^{7}$ .

Favorable cases= no of choices after tossing 7 times all the faces appear at least once.

So The possible arrangement is  $1,2,3,4,5,6,x$ . [ $x$ can be anything between $1-6$]

Favorable choices= $\large \binom{6}{1}*\frac{7!}{2!}$  [ choosing $x$ in $\large \binom{6}{1}$ ways then permute then in $\large \frac{7!}{2!}$]

Probability = $\large \frac{6*7!}{2! *6^{7}}$=$\large \frac{7!}{2! *6^{6}}$ -----[desired answer]
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My thought process is 

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Doubt :- probability β†’ $\large \frac{1}{6^{6}}=\frac{6}{6^{7}}$

as from the problem sample space is $\large {6^{7}}$ so favorable cases is $6$ .

But here favorable cases

${1,2,3,4,5,6,6}$

${1,2,3,4,5,6,5}$

${1,2,3,4,5,6,4}$

${1,2,3,4,5,6,3}$

${1,2,3,4,5,6,2}$

${1,2,3,4,5,6,1}$

${2,3,4,5,6,1,1}$

so on

so 6 favorable cases is not correct .

Check the calculation once.
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Yeah i forgot to do the permutations of the outcomes  my bad πŸ˜…

thnx for pointing it out
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