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I keep on having problems with the same kind of questions involving combinatorics. This is one such problem where I am stuck.

A bag contains 9 discs labelled 1 to 9. Andy chooses 4 discs at random, without replacement.

Find the probability that:

a) the 4 digits include at least 3 odd digits

b) the 4 digits add up to 28

I started by working out that there are 5 odd digits and 4 even digits, and that there are $^9P_4$ different numbers the 4 digits can form. But now I don't know where to go from here. Please also describe a way of going about these kinds of questions. Thanks!

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  • $\begingroup$ You will find some good general advice, techniques, and worked problems by going through the other questions tagged combinatorics. Some of them recommend working on a simpler version of the problem first. How would you answer this one if there were just 3 discs randomly chosen from among 5 discs labeled 1 through 5 and the questions asked for the probability (a) of including at least 2 odd digits and (b) that the 3 digits sum to 9? That's a small enough problem that you can completely enumerate all the possibilities; there are only 10 of them. $\endgroup$
    – whuber
    Jun 1 '15 at 15:56
  • $\begingroup$ @whuber Thanks, are there not 60 possible numbers (instead of 10) formed because it is a permutation; not a combination? $\endgroup$
    – Cobbles
    Jun 1 '15 at 16:04
  • $\begingroup$ Ask yourself this: how would the sequence in which the four discs are chosen affect the events (a) and (b)? Another way to approach this issue is to recognize that the problem statement is ambiguous: it doesn't actually say the discs are chosen sequentially. It asserts only that four were chosen. Maybe they were grabbed all at once. Thus you cannot even uniquely identify a permutation in Andy's sample. Consequently, you shouldn't assume any information about a definite permutation. $\endgroup$
    – whuber
    Jun 1 '15 at 16:07
  • $\begingroup$ @whuber But isn't it a permutation because say one order is 123, another one would be 321 which is a different number? Does whether they were grabbed all at once affect whether its a permutation or a combination? $\endgroup$
    – Cobbles
    Jun 1 '15 at 16:10
  • $\begingroup$ I recommend you focus on solving the problem rather than on trying to shoehorn it into some kind of permutation vs. combination template. The solution to the smaller version is straightforward and takes little time because you can look at every possibility. The purpose of doing that is to see what patterns are emerging and why, then use that understanding to address the original problem. $\endgroup$
    – whuber
    Jun 1 '15 at 16:25
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I've got it

a) One can either get 3 odd numbers or 4; there are 5 odd discs and 4 even discs, so $^5C_3 \times ^4C_1 + ^5C_4 = 45$ divide that by $^9C_4$

b) There are only 2 possibilities because he has not laid them out in order so 9, 8, 7, 4 and 9, 8, 6, 5 are the only ones that add up to 28.

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