 # Remeron

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By: S. Umul, M.A., M.D., M.P.H.

Medical Instructor, CUNY School of Medicine

We can consider all C(7 treatment for depression order 15mg remeron visa, 2) Ч C(4 treatment 5th finger fracture buy remeron line, 2) 2-women­2-men committees and then subtract the forbidden committees that contain both Bagginses medications qhs buy line remeron. The forbidden committees are formed by picking one more woman and one more man to join Mr. When counting the ways to pick elements in a given subset, as a part of a more complex problem, one needs to specify the number of elements in the subset. If the size of the subset can vary, then one must break the problem into subcases so that the size of the subset is a fixed number in each subcase. The mistake of counting the same outcome twice, which arose in part (d) of Example 5, arises in many guises. The Set Composition Principle Suppose a set of distinct objects is being enumerated using the multiplication principle, multiplying the number of ways to form some first part of the set by the number of ways to form a second part (for a given first part). Another way to express this condition is: given any set S thus constructed, one must be able to tell uniquely which elements of S are in the first part of S and which elements are in the second part. In the "committee with at least two women" problem in Example 5(d), the method of choosing two of the seven women first and then picking any remaining two people (women or men) violates the Set Composition Principle because the members of the two parts are not disjoint-any woman could be in either the first or second part. For example, in the committee {W1, W2, W3, M3 } it is impossible to say which two women are chosen in the first part. The scheme of picking two women and then any two remaining people generates the committee {W1, W2, W3, M3 } in C(3, 2) different ways-namely, all compositions of (i) two of the three women W1, W2, W3; with (ii) M3 plus the remaining woman in W1, W2, W3 not chosen in (i). In the following continuation of that example, we consider the constraint of requiring a particular letter to appear somewhere before another letter in the arrangement. How many arrangements have the E somewhere before the M and the three Ss grouped consecutively? The key to the constraint of E being somewhere before the M (not necessarily immediately before the M) is to focus on the pair of positions where E and M will go. Thus, we start by picking which of the two out of the seven positions in an arrangement are where the E and M will go-C(7, 2) = 21 ways-and then we put E in the first of this pair of positions and M in the second one. Now we fill in the five other positions in the arrangement by picking a position for the Y and the T-P(5, 2) = 5 Ч 4 = 20 ways-and then putting the three Ss in the three remaining positions. While it may sound scary to deal with the two constraints at once, it often turns out to be less hard than expected if one handles the constraints in the right order. If we first pick the pair of the positions for the E and the M, things get messy for the consecutivity constraint because the different placements of the E and M will impact differently the positions in the arrangement where there is enough room to place three consecutive Ss. Now we turn to the other constraint and pick the pair of positions, out of the five new positions, for the E and the M-C(5, 2) = 10 ways-with the E going in the first of the two chosen positions. At the last stage, an inspector marks the ovens A (acceptable) or U (unacceptable). How many different sequences of 15 As and Us are possible in which the third U appears as the twelfth letter in the sequence? This problem is a binary sequence problem similar to Example 5 except now the elements are A and U, rather than 0 and 1. If the third U appears at the twelfth letter in the sequence, then the subsequence composed of the first 11 letters must have exactly two Us (and nine As). Following the reasoning in Example 5, there are C(11, 2) = 55 possible sequences for the first 11 letters. The remaining three letters in the sequence can be either A or U -23 = 8 possibilities. Example 7: Probability of Repeated Digits What is the probability that a 4-digit campus telephone number has one or more repeated digits? We break the problem of counting 4-digit phone numbers with repeated digits into four different cases of repetitions: (a) All four digits are the same. First pick which digit appears once-10 choices-then where it occurs in the 4-digit number-four choices-and finally which other digit appears in the other three positions-nine choices. In sum, there are 10 + 360 + 270 + 4320 = 4960 4-digit phone numbers with a repeated digit. One point of caution about cases (c) and (d) where two different digits both occur once or both occur twice. In case (d), we pick the two digits occurring once as an unordered pair in C(10, 2) ways and arrange those digits (and then pick the digit to go in the remaining two positions) rather than pick a first digit, position it, then pick a second digit, position it (and then pick the digit to go in the remaining two positions)-10 Ч 4 Ч 9 Ч 3 Ч 8 ways. In this latter (wrong) approach, we cannot tell for a telephone number such as 2529 whether the 5 was chosen first and put in the second position and then the 9 chosen next and put in the fourth position, or whether the 9 was chosen first and put in the fourth position and then the 5 chosen next and put in the second position. The disjointness requirement of the multiplication principle is being violated and each outcome in case (d) would be counted twice.   