Question

In an examination, the maximum mark for each of the 3 papers is 50 and the maximum mark for the fourth paper is 100. Find the number of ways in which the candidate can score 60% marks in aggregate.
(a) 110550
(b) 110551
(c) 122050
(d) 122051

Answer 1

Hint: Use theory of multinomials and combinations to solve this question.
The multinomial theorem states that:

\[{{\left( {{a}_{1}}+{{a}_{2}}+...+{{a}_{k}} \right)}^{n}}=\sum\limits_{\begin{smallmatrix}
 {{j}_{1}},{{j}_{2}},...{{j}_{k}} \\

 0\le {{j}_{i}}\le n\text{ for each i}

 \\

 \text{and }{{\text{j}}_{1}}+...+{{j}_{k}}=n

\end{smallmatrix}}{\dfrac{n!}{{{j}_{1}}!.{{j}_{2}}!...{{j}_{k}}!}.a_{1}^{{{j}_{1}}}.a_{2}^{{{j}_{2}}}..

.a_{k}^{{{j}_{k}}}}\]

Complete step by step answer:
First we will find total marks,

Total marks = 50 + 50 + 50 + 100 = 250

Asked for 60% of total aggregate,

60% of total = 60% of 250 = 150.

So we need the number of ways a student can score 150 marks out of 250.

Let us assume the following:

Marks obtained in the first test are x.

Marks obtained in the second test are y.

Marks obtained in the third test are z.

Marks obtained in the fourth test are w.

So we require the total to be 150,

x + y + z + w = 150.

Where,

\[0\le x,y,z\le 50\text{ };\text{ 0}\le \text{w}\le \text{100}\]

Now by applying multinomial theorem, we get:

As the first 3 tests have 50 as maximum and next test has 100 as maximum we need to write expression as follows

\[=\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{50}} \right).\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{50}} \right).\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{50}} \right).\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{100}} \right)\]

By multinomial theorem we can say Number of ways a student can score 150 is Coefficient of \[{{x}^{150}}\] in the expression:

\[{{\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{50}} \right)}^{3}}.\left( 1+{{x}^{1}}+{{x}^{2}}.......+{{x}^{100}} \right).....\left( 1 \right)\]

So our required answer would be the Coefficient of \[{{x}^{150}}\] in equation (1).

By using sum of Geometric progression:

\[a+ar+a{{r}^{2}}+....+a{{r}^{n-1}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\]

Here the expression changes to:

\[={{\left( \dfrac{1-{{x}^{51}}}{1-x} \right)}^{3}}\left( \dfrac{1-{{x}^{101}}}{1-x} \right)\]

By expanding the term using the algebraic identity:

\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}\]

By using this, we get:

\[=\dfrac{\left( 1-{{x}^{153}}-3{{x}^{51}}+3{{x}^{102}} \right)\left( 1-{{x}^{101}} \right)}{{{\left( 1-x \right)}^{4}}}\]

By multiplying and dividing with\[{{\left( 1-x \right)}^{-4}}\], we get:

\[=\dfrac{\left( 1-{{x}^{153}}-3{{x}^{51}}+3{{x}^{102}} \right)\left( 1-{{x}^{101}} \right){{\left( 1-x \right)}^{-4}}}{{{\left( 1-x \right)}^{4}}{{\left( 1-x \right)}^{-4}}}=\left( 1-{{x}^{153}}-3{{x}^{51}}+3{{x}^{102}} \right)\left( 1-{{x}^{101}} \right){{\left( 1-x \right)}^{-4}}\]

By multiplying the first two terms using distributive law, we get:

(a + b) . (c) = a.c + b.c

\[=\left(

1-{{x}^{153}}-3{{x}^{51}}+3{{x}^{102}}-{{x}^{101}}-{{x}^{254}}-3{{x}^{152}}-2{{x}^{203}} \right){{\left( 1-x \right)}^{-4}}.....\left( 2 \right)\]

We need the value of \[{{\left( 1-x \right)}^{-4}}\]to proceed further.

Use the binomial theorem to find the value:

\[{{\left( 1-x \right)}^{-n}}=1+nx+\dfrac{n\left( n+1 \right)}{2!}{{x}^{2}}+\dfrac{n\left( n+1 \right)\left( n+2 \right)}{3!}{{x}^{3}}......\]


Finding value of \[{{\left( 1-x \right)}^{-4}}\]:

By substituting n = 4 in above equation, we get:

\[{{\left( 1-x \right)}^{-4}}=1+{}^{4}{{C}_{1}}x+{}^{5}{{C}_{2}}{{x}^{2}}+......\]

By substituting this into equation (2), we get:

\[=\left(

1-{{x}^{153}}-3{{x}^{51}}+3{{x}^{102}}-{{x}^{101}}-{{x}^{254}}-3{{x}^{152}}-2{{x}^{203}}

\right)\left( 1+{}^{4}{{C}_{1}}x+{}^{5}{{C}_{2}}{{x}^{2}}+...... \right)\]

By binomial theorem:

Coefficient of \[{{x}^{r}}\] in \[{{\left( 1-x \right)}^{-n}}\] is \[{}^{n+r-1}{{C}_{r}}\].

Now we need all coefficients of \[{{x}^{150}}\] and add them.

Some of the terms in the first bracket are itself > 150 so they cannot give \[{{x}^{150}}\].

So the possible cases where we can get \[{{x}^{150}}\]are:

Case-1

As there is 1 in first bracket we need the coefficient of \[{{x}^{150}}\]in second bracket
From above equation, we get:

Coefficient of \[{{x}^{150}}\]in second bracket =\[{}^{153}{{C}_{150}}\]…..(3)

Case-2

As there is \[-3{{x}^{51}}\]in first bracket we need the coefficient of \[{{x}^{99}}\]in second bracket

From above equation, we get:

Coefficient of \[{{x}^{99}}\]in second bracket =\[{}^{102}{{C}_{99}}\]…..(4)

Case-3

As there is \[3{{x}^{102}}\]in first bracket we need the coefficient of \[{{x}^{48}}\]in second bracket

From above equation, we get:

Coefficient of \[{{x}^{48}}\]in second bracket = \[{}^{51}{{C}_{48}}\]…..(5)

Case-4

As there is \[-{{x}^{101}}\]in first bracket we need the coefficient of \[{{x}^{49}}\]in second bracket

From above equation, we get:

Coefficient of \[{{x}^{49}}\]in second bracket = \[{}^{52}{{C}_{49}}\]…..(6)

By keeping algebraic coefficients in mind, we get:

Coefficient of \[{{x}^{150}}\] = equation (3) – (3\[\times \]equation (4)) + (3 \[\times \]equation (5)) – equation (6)

= \[{}^{153}{{C}_{150}}\] - 3. \[{}^{102}{{C}_{99}}\] + 3. \[{}^{51}{{C}_{48}}\] - \[{}^{52}{{C}_{49}}\]

\[=\dfrac{153!}{150!3!}-\left( 3.\dfrac{102!}{99!3!} \right)+\left( 3.\dfrac{51!}{48!3!} \right)-\dfrac{52!}{49!3!}\]

=585276 - 3(171700) + 3(20825) - 22100

= 110551

Therefore the number of ways possible to get 60% aggregate is 110551 ways.

Option (b) is correct


Note: Be careful while calculating the coefficients by binomial and multinomial, if you forgot to subtract 1 in the formula you may lead to the wrong answer.
Expansion done by binomial expansion is a crucial step. Do it carefully.