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Finding Number of cases? Suppose we have 40 cards numbered 1 through 40 and we want to pick 6 of them one at a time such that order of picking card “does matter”.

A. probability of numbers 1 and 2 are not among those cards?

B. probability of at least numbers one of numbers 10, 11, 12 are among those cards?

C. probability of at least numbers two of numbers 20, 21, 22 are among those cards?

Please correct me . Thanks

a) P(38,6)/P(40,6)

b) 1- [p(37,6)/p(40,6)]

I have no idea for part c

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Your expressions for a) and b) are correct. Instead of the permutation symbol that you called $P(n,6)$, you could have used $\dbinom{n}{6}$. Doing that divides the numerator and denominator by $6!$, so it makes no difference to the answer. But it's a good thing to get used to. In the future, you will often find yourself using $\dbinom{n}{k}$, while $P(n,k)$ will be less useful.

There are various approaches to c). The most simple-minded way to count is to first count the number of choices that have all of $20$, $21$, and $22$. To count these, note that we need to choose $3$ items from the remaining $37$. This can be done in $\dbinom{37}{3}$ ways. But there are $6!$ orders in which the numbers could be chosen, so the total count (if later we will use $P(40,6)$ in the denominator) is $6!\dbinom{37}{3}$.

Now count the hands that have exactly $2$ of $20$, $21$, $22$. We can decide which $2$ in $\dbinom{3}{2}$ ways, and for each way we can select the $4$ outsiders that will join them in $\dbinom{37}{4}$ ways. Finally, multiply by $6!$ in order to scramble them in all possible orders.

But I prefer not to use $P(40,6)$ in the denominator, and instead use $\dbinom{40}{6}$. Then the numerator is $\dbinom{37}{3}+\dbinom{3}{2}\dbinom{37}{4}$.

There are other ways to count. In particular, it will after a while be important to be able to handle the method of Inclusion/Exclusion described in the answer by Ross Millikan.

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A and B are fine. Note that it isn't important whether the order matters or not. If you say it doesn't matter, all your P's become C's, but the answer is the same. For C, you need the inclusion-exclusion principle. You can calculate the probability that 20 and 21 are both among the cards, then 21 and 22, then 20 and 22, and add them up. The problem is you have triple counted the cases where all three are included. So you need to subtract twice the chance that all three are included and you have your answer.

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