The explanation says that A, B, and C cannot be all equal. Why is it possible to have A=B≠C?

  • M1★ M3★ M4 M5★

    The explanation says that \(A, B, C\) cannot be all equal. Why is it possible to have \(A=B \neq C?\) This also includes \(A=C \neq B\) and \(B=C \neq A.\)

    How do you do it if that is possible?

  • ADMIN M0★ M1 M5

    @spaceblastxy1428 Thank you for asking and for thinking! The question is saying that we should not count the case where \( A = B = C,\) and this happens when there are \(4\) of each grade.

    You could have, for example, have \(2 A\)'s, \(2 B\)'s, and \(8 C\)'s, giving \(A = B \neq C.\) Or you could have \(1 B, 1C,\) and \(10 A\)'s, giving \(B = C \neq A.\)

    If you look at the question as a whole, we need not worry about the particular ways to assign the grades; we can simply count the ways to construct a diagram of \(11\) identical stars and \(2\) identical bars. There is a one-to-one correspondence between a formation of these stars and bars and a corresponding way to assign the grades.

    M3W2D8-y-part-2-solution2.png

    Someone else recently asked about this question, and you're welcome to take a look at that explanation here! 🙂

  • M1★ M3★ M4 M5★

    @debbie Is there a way to do it if \(A=B \neq C, A=C \neq B, \text{and } B=C \neq A\) are impossible?

  • ADMIN M0★ M1 M5

    @spaceblastxy1428 Are you asking if we could answer a different but similar question, which is that we want the ways to create grade curves where no two batches of \(A\) grades, \(B\) grades or \(C\) grades have the same number of students in them?

    This is definitely possible! It's relatively easy to count the ways to have exactly two of the grade batches have an equal number of people. Here because of symmetry, the number of ways to have \( A = B \neq C \) equals the number of ways to have \(B = C \neq A\) and also equals the number of ways to have \( A = C \neq B, \) so let's just consider the ways for one of these, and multiply by \(3\) to find the total ways for \( A = B \neq C, B = C \neq A, \) or \( A = C \neq B.\)

    The sum of the two equal batches will be an even number, so the non-equal batch will have an even number of students in it. The different ways to have students in the non-equal batch are:

    $$2 \\ 4\\ 6\\ 8\\ 10\\ $$

    since we cannot have an empty batch. This is \(5\) ways, so the fast way is to multiply \(5\) by \(3\) since there are three ways ( \( A = B, B = C, \text{ or } A = C.)\)

    This gives us \(15\) ways that we don't want. However, we counted the way \( A = B = C = 4\) three times, so we must subtract \(2\) to get \(13\) ways that we don't want.

    There were \(\binom{11}{2} = 55 \) ways to arrange the \(9\) stars and \(2\) bars, so removing the \(13\) ways that we don't want ( corresponding to \(A = B, B = C, \text{ or } A = C) \) we get a total of

    $$ 55 - 13 \text{ ways } = \boxed{42 \text{ ways to have a grade curve if there are different numbers of students getting each grade}} $$

    I hope this helps, and thanks so much for asking! 🙂