Question

A test has 4 parts; the first 3 parts carry 10 marks each and the 4th part carries 15 marks. Assuming that marks are not given in the form $\dfrac{p}{q}$ (p < q, p, q ∈ I), $q\ne 0$, then the number of ways in which the student can get 30 marks out of 45 is
(a) 5300
(b) 711
(c) 6011
(d) None of these

Answer 1

To get the number of ways the student can get 30 marks out of 45, we will make use of binomial theorem. We know that the probability of marking a question correct is equal to marking another question correct and is independent of it. This means, students can have 0 mark from any one of the parts and rest of the marks from others. Keeping this concept in mind, we will find the number of ways.

Complete step by step answer:
The first three parts have maximum 10 marks each and the fourth part has 15 marks.
We can safely deduce that students can score any number of marks from any parts but the total marks should be 30 out of 45.
So, from part 1, the student can score 0 marks 1 marks or 2 marks or 3 marks and so on up to 10 marks.
Thus, binomial distribution for part 1 = ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}}$
Similarly, we can write the conditions for part 2 and part 3 are the same as that of part 1.
Thus, binomial distribution for part 2 = ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}}$
And binomial distribution for part 3 = ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}}$
Now, in the fourth part, the maximum marks that can be scored by the student is 15. This means, he can score 0 marks, or 1 mark, or 2 marks and so on up to 15 marks from this part.
Thus, binomial distribution for part 4 = ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{15}}$
Since, it is understandable to us that all these parts of tests are happening simultaneously and student has to score marks collectively, the binomial distribution for scoring collectively from 4 tests will be
$\Rightarrow \left( {{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}} \right)\left( {{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}} \right)\left( {{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}} \right)\left( {{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{15}} \right)$
Now, if the student has to score 1 mark, he can score it from any of the test, so, the number of ways he can score 1 mark from all the tests will be the coefficient of ${{x}^{1}}$ from the product of the binomial.
Similarly, if we want to know the number of ways, he can score 30 marks, we need to find the coefficient of ${{x}^{30}}$ in the resultant binomial.
$\Rightarrow {{\left( 1+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}} \right)}^{3}}\left( 1+{{x}^{1}}+{{x}^{2}}+....+{{x}^{15}} \right)$
Now, we know from the concepts of binomial expansion theorem that ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{10}}=\left( \dfrac{{{x}^{11}}-1}{x-1} \right)$
Similarly, ${{x}^{0}}+{{x}^{1}}+{{x}^{2}}+....+{{x}^{15}}=\left( \dfrac{{{x}^{16}}-1}{x-1} \right)$
Thus, we have to find coefficient of ${{x}^{30}}$ in the expansion of ${{\left( {{x}^{11}}-1 \right)}^{3}}\left( {{x}^{16}}-1 \right){{\left( x-1 \right)}^{-4}}$.
Expansion of ${{\left( {{x}^{11}}-1 \right)}^{3}}={{x}^{33}}-3{{x}^{22}}+3{{x}^{11}}-1$
Expansion of ${{\left( x-1 \right)}^{4}}=\dfrac{1}{{{x}^{4}}}+\dfrac{4}{{{x}^{5}}}+\dfrac{20}{{{x}^{7}}}+\dfrac{35}{{{x}^{8}}}+\dfrac{56}{{{x}^{9}}}+\dfrac{84}{{{x}^{10}}}+\dfrac{120}{{{x}^{11}}}+\dfrac{165}{{{x}^{12}}}+\dfrac{220}{{{x}^{13}}}+\dfrac{286}{{{x}^{14}}}+\dfrac{364}{{{x}^{15}}}+\dfrac{455}{{{x}^{16}}}+\dfrac{560}{{{x}^{17}}}+\dfrac{680}{{{x}^{18}}}+\dfrac{816}{{{x}^{19}}}+\dfrac{969}{{{x}^{20}}}+....$
Thus, we need to find coefficient of ${{x}^{30}}$ in $\left( {{x}^{33}}-3{{x}^{22}}+3{{x}^{11}}-1 \right)\left( {{x}^{16}}-1 \right)\left( \dfrac{1}{{{x}^{4}}}+\dfrac{4}{{{x}^{5}}}+\dfrac{20}{{{x}^{7}}}+\dfrac{35}{{{x}^{8}}}+\dfrac{56}{{{x}^{9}}}+\dfrac{84}{{{x}^{10}}}+\dfrac{120}{{{x}^{11}}}+\dfrac{165}{{{x}^{12}}}+\dfrac{220}{{{x}^{13}}}+\dfrac{286}{{{x}^{14}}}+\dfrac{364}{{{x}^{15}}}+\dfrac{455}{{{x}^{16}}}+\dfrac{560}{{{x}^{17}}}+\dfrac{680}{{{x}^{18}}}+\dfrac{816}{{{x}^{19}}}+\dfrac{969}{{{x}^{20}}}+.... \right)$
Thus, the coefficient of ${{x}^{30}}$ is 816 – 3(35) = 711.

So, the correct answer is “Option B”.

Note: Binomial expansion is used when a binomial is multiplied by itself n number of times. If (a + b) is binomial, then ${{\left( a+b \right)}^{n}}$ $=\left( \begin{aligned} & n \\\ & 0 \\\ \end{aligned} \right)$ ${{a}^{n}}{{b}^{0}}+$ $\left( \begin{aligned} & n \\\ & 1 \\\ \end{aligned} \right)$ ${{a}^{n-1}}{{b}^{1}}+\left( \begin{aligned} & n \\\ & 2 \\\ \end{aligned} \right){{a}^{n-2}}{{b}^{2}}+...+$ $\left( \begin{aligned} & n \\\ & n \\\ \end{aligned} \right){{a}^{0}}{{b}^{n}}$
Where $\left( \begin{aligned} & n \\\ & r \\\ \end{aligned} \right){{=}^{n}}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$