Number of ways to divide n identical objects into groups

Question

Question 1: If a student can score a maximum of $100$ marks in three subjects p,c,m, then find the number of ways in which he can score a total of $190$ marks while getting at least $50$ in each subject.

Procedure:

$$x_1 + x_2 +x_3 = 190$$

$x_1 \geq 50, x_2 \geq 50, x_3 \geq 50$

which implies

$$x_1+x_2+x_3 = 40$$

$x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$

Number of ways $= \binom{40 + 3 - 1}{3-1} = \binom{42}{2}$

I understand this is the right answer.

Question 2: If a student can score maximum of $100$ marks in three subjects p,c,m, then find the number of ways in which he can score a total of $230$ marks while getting at least $50$ in each subject.

The formula used above gives $\binom{82}{2}$.

I understand this is not the right answer.

I also do understand that the above formula comes from multinomial theorem. I wanted to check why the formula works for question 1 and doesn't work for question 2.

Answer 1

You took the minimum scores into account but did not do the same for the maximum scores.

If a student can score a maximum of $100$ marks in each of three subjects, in how ways can he score a total of $190$ marks while getting a score of at least $50$ in each subject.

Let $x_i$ be the score in the $i$th subject. Then $$x_1 + x_2 + x_3 = 190 \tag{1}$$ subject to the restrictions that $50 \leq x_i \leq 100$ for $1 \leq i \leq 3$.

Let $y_i = x_i - 50$, $1 \leq i \leq 3$. Then each $y_i$ is a nonnegative integer satisfying $0 \leq y_i \leq 50$. Substituting $y_i + 50$ for $x_i$, $1 \leq i \leq 3$, in equation 1 yields \begin{align*} y_1 + 50 + y_2 + 50 + y_3 + 50 & = 190\\ y_1 + y_2 + y_3 & = 40 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers, which, as you observed, as $$\binom{42}{2}$$ solutions, none of which violate the restrictions that $y_i \leq 50$, $1 \leq i \leq 3$. Therefore, your solution is correct.

If a student can score a maximum of $100$ marks in each of three subjects, in how ways can he score a total of $230$ marks while getting a score of at least $50$ in each subject.

Let $x_i$ be the score in the $i$th subject. Then $$x_1 + x_2 + x_3 = 230 \tag{3}$$ subject to the restrictions that $50 \leq x_i \leq 100$ for $1 \leq i \leq 3$.

Let $y_i = x_i - 50$, $1 \leq i \leq 3$. Then each $y_i$ is a nonnegative integer satisfying $0 \leq y_i \leq 50$. Substituting $y_i + 50$ for $x_i$, $1 \leq i \leq 3$, in equation 3 yields \begin{align*} y_1 + 50 + y_2 + 50 + y_3 + 50 & = 230\\ y_1 + y_2 + y_3 & = 80 \tag{4} \end{align*} Equation 2 is an equation in the nonnegative integers, which, as you observed, as $$\binom{82}{2}$$ solutions. However, in this case, it is possible that one of the restrictions $y_i \leq 50$, $1 \leq i \leq 3$ is violated. It is not possible to violate more than one of the restrictions simultaneously since $2 \cdot 51 = 102 > 80$.

There are three ways to choose which variable violates the restriction that $y_i \leq 50$. Suppose it is $y_1$. Then $y_1 \geq 51$. Let $z_1 = y_1 - 51$. Then $z_1$ is a nonnegative integer. Let $z_2 = y_2$; let $z_3 = y_3$. Substituting $z_1 + 50$ for $y_1$, $z_2$ for $y_2$, and $z_3$ for $y_3$ in equation 4 yields \begin{align*} z_1 + 51 + z_2 + z_3 & = 80\\ z_1 + z_2 + z_3 & = 29 \tag{5} \end{align*} Equation 5 is an equation in the nonnegative integers with $$\binom{29 + 3 - 1}{3 - 1} = \binom{31}{2}$$ solutions.

Since there are three ways to select the variable that violates the restriction that $y_i \leq 50$, $1 \leq i \leq 3$, there are $$\binom{3}{1}\binom{31}{2}$$ solutions of equation 4 that violate one of the restrictions.

Therefore, there are $$\binom{82}{2} - \binom{3}{1}\binom{31}{2} = 1926$$ ways for the student to obtain a total of $230$ marks with a score of at least $50$ in each subject, as Ross Millikan found using a different method.

Answer 2

For Question 1:

You can instantly ignore cases where any of the three scores are less than 50. So assume you're getting at least 50 in each test. Then the additional score needed is $40$ ($= 190 - 3 \cdot 50$). Sum over these conditions with the total being $40$:

$$\sum\limits_{p=0}^{40} \sum\limits_{c=0}^{40-p} \sum\limits_{m=0}^{40-c-p} 1 = 12341$$

I'll let you apply the same logic to Question 2.

Answer 3

In the first, you can just subtract $50$ from each score and find the ways to sum three numbers (including $0$) up to $40$ The stars and bars calculation for this is $42 \choose 2$ as you say.

The difference in the second is that you need $80$ more points but no one test can give you more that $50$. $82 \choose 2$ is the number of ways to add up three numbers (including $0$) to get $80$ if any of them can give you the full $80$. Now you need to find the number of ways to add up three numbers in the range $[0,50]$ to get $80$.

I don't have a neat way to count the second. If you get $0$ points on the first subject there are $21$ possibilities. If you get $30$ points on the first subject there are $51$ possibilities. If you get $50$ points on the first there are $31$ possibilities. It is linear between these, so the number of possibilities is $$\sum_{i=0}^{30}(21+i)+\sum_{j=31}^{50}(81-j)=21\cdot31+\frac 12\cdot 30 \cdot 31+81\cdot 20-\frac 12\cdot 81\cdot 20=1926$$