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We roll $2$ different die $n$ times, with respect to order of the die and the order for $n$ times. What is the number of solutions where all the following doubles $(1,1),(2,2),...,(6,6)$ exist?

Start by calculating the overall possible combinations of throws: $$(6^2)^n = 36^n$$

The possible combinations of getting at-least one double is $$2^1 \binom{6}{1}36^{n-1}$$

Overall for getting at-least $i$ doubles we get $$2^i \binom{6}{i}36^{n-i}$$

I've been suggested to use inclusion-exclusion but I fail to understand why is it needed here. Why isn't the number of solutions simply $2^6 \binom{6}{6}36^{n-6}$?

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  • Inclusion-exclusion is the way to go. As for your attempt, the $2^6$ term is entirely out of place... surely you meant to have $6!$ or a $\dfrac{6!}{(6-i)!}$ instead. Next, you don't have a $\binom{n}{i}$ term which commonly appears in mistakes like yours... that makes it easier to explain your answer. You effectively asked how many ways very specifically the first $i$ rolls were doubles and the remaining rolls were whatever, including possibly more doubles. – JMoravitz May 12 '20 at 14:45
  • Recall, it is possible for there to be more than one occurrences of each desired double. Further, there is no requirement that the doubles occur very specifically at the start of the sequence, they could start later in the sequence. We don't want to accidentally overcount however, so care needs to be taken. – JMoravitz May 12 '20 at 14:47
  • This is a truncated coupon collector's problem – you're looking for the probability to have $6$ desired coupons out of $36$ different coupons after $n$ draws. This is answered here and here. I'm not closing this question as a duplicate because you didn't ask about the right answer but about why your wrong answer is wrong. – joriki May 12 '20 at 14:51

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