I have to solve next task: form numbers 1, 2 and 3 is randomly made six digit number. What is probability to number 1 to be only one time, number 2 to be two times and number 3 to be three times?
Six digit number probability
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Total number of 6 digit number made using $1,2,3$ only $=3^6\,\,\,\,\,\,\,\,\,(1)$
Number of numbers with one 1, two 2s and three 3s $=\frac{6!}{2!3!}=60\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$
Probablity = $60/3^6\approx8.23\%$
Explanation for $(1)$:
Consider $6$ blank spaces. In each of those spaces we can fill $1,2$ or $3$ thus each has three choices and hence total is $3\times3\times3\times3\times3\times3=3^6$
Explanation for $(2)$:
Consider the number $122333$
We can rearrange the digits in $6!$ ways. But since $2$ and $3$ are repeated, we must divide the number of times they are repeated since we are over counting 2 times and 6 times repectively. You may also look at this http://www.mathwarehouse.com/probability/permutations-repeated-items.php
All possibilities: Every 6-digit number that contains $1$, $2$ or $3$:
Overall: $3 \times 3 \times \dots \times 3 = 3^6 = 729$ possibilities.
Beneficial possibilities: Since $1+2+3=6$, you're technically ordering these digits: $$1,\ 2,\ 2,\ 3,\ 3,\ 3$$ The ways you can order $6$ elements is $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.
However, some of the elements are indistinguishable, so we need to reduce the ways you can order them by the ways you can swap out $2$'s for other $2$'s and $3$'s for other $3$'s. You can swap them out $2!$, and $3!$ ways.
So all the number of distinguishable ways to order these values is: $$\frac{6!}{2! \times 3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} = 60.$$
Probability: $$\Bbb{P}(\text{The number contains 1 digit-1, 2 digit-2's and 3 digit-3's}) = \frac{60}{729} \approx 0.0823 = 8.23\%$$