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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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?

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{The AMS MOCK CAT test CATALYST 19 consists of four sections .}\hspace{.33em}\\~\\ & \normalsize \text{ Each section has a maximum marks of 45 marks. Find the number}\hspace{.33em}\\~\\ & \normalsize \text{ of ways in which a student can qualify in the AMS MOCK CAT if }\hspace{.33em}\\~\\ & \normalsize \text{the qualifying mark is 90? }\hspace{.33em}\\~\\ & a.)\ 36546 \hspace{.33em}\\~\\ & b.)\ 6296 \hspace{.33em}\\~\\ & c.)\ 64906 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{None of these} \hspace{.33em}\\~\\ \end{align}}\)
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 \(0 \le x_i\le 45 \)
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? :)

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is this correct \(\large \color{black}{\begin{align} & \dbinom{180}{90} \hspace{.33em}\\~\\ \end{align}}\)
Is that your final answer? Because that's a huuuuuuge number.
One question: does the distribution of marks matter within each section?
I think not... that would make the question way too heavy.
answer given in book is \(C.)\)
I think the distribution of marks doesn't matter within each section
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?
No
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 \((1+x+x^2+x^3+x^4)^3\)
I have leant stars ans bars , multinomial is not in my sylabus
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.
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?
Basically I'm asking how you've been taught to do that.
i m not sure that stars and bars can be used here, i just thought that it can be used in this situation
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:\[\text{No. of solutions }=\text{Coeff. of }x^2 \text{ in }(1+x+x^2+x^3 + \cdots)(1+x+x^2+x^3+\cdots) \]
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:\[1 \times x^2 = x^2\]
Similarly, \(1+1=2\) corresponds to:\[x \times x = x^2\]And \(2+0 =2\) corresponds to:\[x^2\times 1 = x^2\]
Which is why the coefficient of \(x^2\) tells you the number of solutions, right? :)
i need to go ,bcuz battry is down, i will read ur expln later
Now\[(1+x+x^2 + \cdots ) (1 + x + x^2 + \cdots)\]\[= \frac{1}{1-x}\cdot \frac{1}{1-x}\]\[= \frac{1}{(1-x)^2}\]\[= 1 + 2x + 3x^2 + 4x^3 + \cdots\]
Therefore, the number of solutions is \(3\).
  • phi
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.
I have found some where this but i cant understand this and not sur if it is correct \(\large \color{black}{\begin{align} & \dbinom{93}{3} - 4\times \dbinom{47}{3} \hspace{.33em}\\~\\ \end{align}}\)
it equals 64906
  • phi
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.
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
  • phi
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)
Yeah, exactly.
  • phi
If the question is how many ways to get exactly 90 correct, then http://mathworld.wolfram.com/Dice.html appears to show an approach
that wolfram page is too technical for noob like me
i think the answer should be none of these
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....\[\binom{93}{3}-4\binom{47}{3}\]
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. :)
  • phi
if the book says the answer is 64906, then the question is worded incorrectly. (not the first time)
  • phi
notice that 93C3 is related to 90 stars and 3 bars problem
yea i dont trust the book
93C3= number of non -negative solns of a+b+c+d=90
  • phi
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 ?
i have studied a+b+c+d=90 kinds of problems its just a coincidence i related to it
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.
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:\[61 + 14 + 10 + 5 = 90\]\[15+60+10+5=90\]\[15+14+56+5=90\]\[15+14+10+51=90\]
So if the former (modified) equation has \(k\) solutions, then it should be clear that the original equation has \(4k\) solutions where \(x_i \ge 45\).
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.
That's a nice solution but if you're lazy you can just use the multinomial method if that's what they call it.
ok thnks

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