Combinatorical Drawing

If there is a pool of 2 Ms, 2 As, 2 Ts, 2Hs, and 1 X, Y, Z, how would you find the probability that if you drew 7 letter cards that you could spell Math from what you drew?


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Mathe Mathe
  • You have given the probability of drawing a M, followed by an A, followed by a T, followed by an H. I'm looking for a technique that likely uses combinations and inclusion/exclusion to compute the probability. Something like: total of 7 letter combinations with at least 1 M, A, T, and H over 11 choose 7.

    • Mathe Mathe

      I'm afraid I don't understand your question, but I think it's different from what is stated in the problem.

  • So there are 11 letters and you draws 7 of them. What is the probability that you can spell Math from what you drew?

    • Mathe Mathe

      I added a new response in the case that the order of the picks is not relevant.

  • On second thought, I think you are overcounting. If you drew a D1 in the 2C1 you could draw a D2 in the 7C3, which is fine, but you could also draw a D2 in the 2C1 and a D1 in the 7C3, which is just a reordering of the prior event.

    • Mathe Mathe

      What do you mean by a D1 and a D2? I think you mean an 'M', for example. I must draw an 'M' at some point. The first M I draw is counted in the 2C1 factor. A second M would be counted in the 7C3.

  • Yes, meant M. I contend that the numerator, which represents the number of ways to have M A T H and 3 other letters is too large. The reason for this is, because, say you let M1 a 2C1, and M2 be in the 7C3. This is a valid draw. However, if in the future, if you have an M2 in the 2C1, and a M1 in the 7C3, with the other 5 characters the same, this event would be counted again with the method you presented.

    • Mathe Mathe

      It would also be counted twice in the denominator.

  • It wouldn't be counted twice in the denominator because the denominator is counting unordered events. If you did a permutation, then sure, but the numerator would need to be adjusted.

The answer is accepted.
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