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mathmath333

  • one year ago

The AMS MOCK CAT test CATALYST 19 consists of four sections . Each section has a maximum marks of 45 marks. Find the number of ways in which a student can qualify in the AMS MOCK CAT if the qualifying mark is 90?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{The AMS MOCK CAT test CATALYST 19 consists of four sections .}\hspace{.33em}\\~\\ & \normalsize \text{ Each section has a maximum marks of 45 marks. Find the number}\hspace{.33em}\\~\\ & \normalsize \text{ of ways in which a student can qualify in the AMS MOCK CAT if }\hspace{.33em}\\~\\ & \normalsize \text{the qualifying mark is 90? }\hspace{.33em}\\~\\ & a.)\ 36546 \hspace{.33em}\\~\\ & b.)\ 6296 \hspace{.33em}\\~\\ & c.)\ 64906 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{None of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. ParthKohli
    • one year ago
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    OK, I see. Let's say that the student gets \(x_1\) on section one, \(x_2\) on section two, and so on. We are told that,\[180 \ge x_1+x_2+x_3+x_4 \ge 90\]constrained to \(0 \le x_i\le 45 \)

  3. ParthKohli
    • one year ago
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    Given \(0 \le x_i \le 45\), first, find the number of solutions to the inequality:\[x_1 + x_2 + x_3 + x_4 \le 180\]Then find the number of solutions to the inequality:\[x_1+x_2 + x_3 + x_4 \le 90\]And finally subtract them. Do you know how to do that? :)

  4. mathmath333
    • one year ago
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    is this correct \(\large \color{black}{\begin{align} & \dbinom{180}{90} \hspace{.33em}\\~\\ \end{align}}\)

  5. ParthKohli
    • one year ago
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    Is that your final answer? Because that's a huuuuuuge number.

  6. ParthKohli
    • one year ago
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    One question: does the distribution of marks matter within each section?

  7. ParthKohli
    • one year ago
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    I think not... that would make the question way too heavy.

  8. mathmath333
    • one year ago
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    answer given in book is \(C.)\)

  9. mathmath333
    • one year ago
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    I think the distribution of marks doesn't matter within each section

  10. ParthKohli
    • one year ago
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    Good. To find the number of solutions to\[x_1 + x_2 + x_3 + x_4 \le 90\]introduce a dummy variable \(u\).\[x_1+x_2+x_3+x_4+u = 90\]Now do you know how to use multinomial theorem?

  11. mathmath333
    • one year ago
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    No

  12. ParthKohli
    • one year ago
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    Wait, so like you haven't learned that yet? The number of solutions to, say, \(x_1 + x_2 + x_3 = 8\) given that \(0 \le x_i \le 4\) is the coefficient of \(x^8\) in \((1+x+x^2+x^3+x^4)^3\)

  13. mathmath333
    • one year ago
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    I have leant stars ans bars , multinomial is not in my sylabus

  14. ParthKohli
    • one year ago
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    OK, then leave that.\[x_1 + x_2 + x_3+x_4=90\]\[x_1+x_2+x_3+x_4 = 91\]\[\vdots\]\[x_1+x_2+x_3+x_4 = 180\]I'm not sure how you handle constraints just using stars-and-bars.

  15. ParthKohli
    • one year ago
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    If you're given like the first equality (=90), how would you just use stars and bars to calculate the number of ways given that each variable can't be more than 45?

  16. ParthKohli
    • one year ago
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    Basically I'm asking how you've been taught to do that.

  17. mathmath333
    • one year ago
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    i m not sure that stars and bars can be used here, i just thought that it can be used in this situation

  18. ParthKohli
    • one year ago
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    If you think about it, multinomial isn't really that hard. Let me explain. If you are to find the number of solutions to a very simple equality, say, \(x_1 + x_2 = 2\) given that both are nonnegative, think about this:\[\text{No. of solutions }=\text{Coeff. of }x^2 \text{ in }(1+x+x^2+x^3 + \cdots)(1+x+x^2+x^3+\cdots) \]

  19. ParthKohli
    • one year ago
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    Coefficient is basically the number of ways you can get to \(x^2\) by picking one term each from both the factors.\[0+2 = 2\]is a solution to the equation and it directly corresponds to this:\[1 \times x^2 = x^2\]

  20. ParthKohli
    • one year ago
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    Similarly, \(1+1=2\) corresponds to:\[x \times x = x^2\]And \(2+0 =2\) corresponds to:\[x^2\times 1 = x^2\]

  21. ParthKohli
    • one year ago
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    Which is why the coefficient of \(x^2\) tells you the number of solutions, right? :)

  22. mathmath333
    • one year ago
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    i need to go ,bcuz battry is down, i will read ur expln later

  23. ParthKohli
    • one year ago
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    Now\[(1+x+x^2 + \cdots ) (1 + x + x^2 + \cdots)\]\[= \frac{1}{1-x}\cdot \frac{1}{1-x}\]\[= \frac{1}{(1-x)^2}\]\[= 1 + 2x + 3x^2 + 4x^3 + \cdots\]

  24. ParthKohli
    • one year ago
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    Therefore, the number of solutions is \(3\).

  25. phi
    • one year ago
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    I have not done problems like this. I will look at this https://probabilityandstats.wordpress.com/2010/04/24/the-multinomial-theorem/ and maybe post later.

  26. mathmath333
    • one year ago
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    I have found some where this but i cant understand this and not sur if it is correct \(\large \color{black}{\begin{align} & \dbinom{93}{3} - 4\times \dbinom{47}{3} \hspace{.33em}\\~\\ \end{align}}\)

  27. mathmath333
    • one year ago
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    it equals 64906

  28. phi
    • one year ago
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    Interesting. It does give 64,906 and you said that is the correct answer. Your textbook should have a hint on what strategy to use to solve this problem. Otherwise, for the moment I'm stuck. But I'm interested in how to do it.

  29. mathmath333
    • one year ago
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    in the text book , no hint is given just the answer c.) is given correct, a+b+c+d = 90 93C3 ways When any of the four >45 a'+b'+c'+d' = 44 Total = 4*47C3 ways Total = 93C3 - 4*47C3 = 129766 - 64860 = 64906 I found this here http://www.pagalguy.com/discussions/official-quant-thread-for-cat-2013-25088456/16390238

  30. phi
    • one year ago
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    That is calculating the number of ways to get exactly 90 correct answers. I interpreted the question of what are the number of ways to get 90 or more correct answers (i.e. qualify means at least 90 correct answers)

  31. ParthKohli
    • one year ago
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    Yeah, exactly.

  32. phi
    • one year ago
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    If the question is how many ways to get exactly 90 correct, then http://mathworld.wolfram.com/Dice.html appears to show an approach

  33. mathmath333
    • one year ago
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    that wolfram page is too technical for noob like me

  34. mathmath333
    • one year ago
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    i think the answer should be none of these

  35. ParthKohli
    • one year ago
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    I also thought it should be an inequality. Anyhow, making sense of why we prefer to use these expressions in \(x\) will help you in the long run.\[(1+x+x^2 + \cdots + x^{45})^4 \]\[= \frac{(1-x^{46})^4}{(1-x)^4}\]\[= (1-x^{46})^4 (1-x)^{-4}\]We need the coefficient of \(x^{90}\) in the above expression.\[= (1 - 4x^{46} + 6x^{92} +\cdots )(1+4x+10x^2 + \cdots )\]Now it looks pretty easy. We need to figure out the coefficient of \(x^{90}\) and \(x^{44}\) in the expansion of \((1-x)^{-4}\). They are given by the following:\[\binom{4+90-1}{4-1} = \binom{93}{3}\]\[\binom{4+44-1}{4-1}=\binom{47}{3}\]Now \(x^{90}\) combines with 1 and \(x^{44}\) combines with \(-4x^{46}\) thus making the final coefficient equal to....\[\binom{93}{3}-4\binom{47}{3}\]

  36. ParthKohli
    • one year ago
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    But this is only the equality case. The actual answer should be way higher than that and by the looks of it, impossible to write down in as few digits as they've used in the choices. :)

  37. phi
    • one year ago
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    if the book says the answer is 64906, then the question is worded incorrectly. (not the first time)

  38. phi
    • one year ago
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    notice that 93C3 is related to 90 stars and 3 bars problem

  39. mathmath333
    • one year ago
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    yea i dont trust the book

  40. mathmath333
    • one year ago
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    93C3= number of non -negative solns of a+b+c+d=90

  41. phi
    • one year ago
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    assuming they want the ways to get to 90 they probably expect you to use stars and bars ? or have you studied a+b+c+d=90 kinds of problems ?

  42. mathmath333
    • one year ago
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    i have studied a+b+c+d=90 kinds of problems its just a coincidence i related to it

  43. ParthKohli
    • one year ago
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    Looks like I understand what your solution means @mathmath333 We want to remove those solutions that contain something higher than 45. It should be clear that one and only one of them can be greater than 45. Think of four invisible numbers for a second, of which you know the sum is 90.\[\boxed{} + \boxed{} + \boxed{} + \boxed{}=90\]You are given that one of them is actually higher than 45. Obviously that can't be. So what you do is subtract 46 from both sides. What effect does this have? It brings the value of the variable that's hiding down from 46 or more to 0 or more. Now we really haven't changed anything, have we?\[\boxed{} + \boxed{} + \boxed{} + \boxed{} - 46 = 90 - 46 = 44\]We sort of have changed a few things. First, there is a loss of information in the second equation about which particular variable had a value higher than 45 - there is no way to know. Any of the four new boxes can be the modified variable. So we take the modified equation and multiply its number of solutions by the number of modified variables we could have.

  44. ParthKohli
    • one year ago
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    Let me illustrate this with an example. To the following solution:\[15 + 14 + 10 + 5 = 44 \]There are four corresponding solutions to the original equation:\[61 + 14 + 10 + 5 = 90\]\[15+60+10+5=90\]\[15+14+56+5=90\]\[15+14+10+51=90\]

  45. ParthKohli
    • one year ago
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    So if the former (modified) equation has \(k\) solutions, then it should be clear that the original equation has \(4k\) solutions where \(x_i \ge 45\).

  46. ParthKohli
    • one year ago
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    And since we're not looking for those, we remove them.\[\binom{93}{3}-4k\]Now \(k\) is just another solution to a stars-and-bars problem with 44 stars and 3 bars.

  47. ParthKohli
    • one year ago
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    That's a nice solution but if you're lazy you can just use the multinomial method if that's what they call it.

  48. mathmath333
    • one year ago
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    ok thnks

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