Excessively Demanding School

For $i = 1, 2, \ldots, 10,$ let $m_i$ be the $i$ th mark that Peter received, and for each $i = 1, 2, \ldots, 9,$ let $a_i = m_{i+1}-m_i$ (the difference between consecutive marks).We note that

1. $25 \le m_1 \le 50.$

2. $a_i \ge 0$ for all $i,$ since Peter's scores never decreased.

3. Since his last score is no more than $50,$ we must have

$\begin{aligned} a_1 + a_2 + \ldots + a_9 &= (m_2-m_1) + (m_3-m_2) + \ldots + (m_{10}-m_9) \\ &= m_{10} - m_1 \\ &\le 50 - m_1. \end{aligned}$

Notice that if we choose $m_1$ and $a_1, \ldots, a_9$ satisfying these conditions, then the rest of the $m_i$ are uniquely determined. Conversely, if we choose the $m_i$ satisfying the conditions, then the $a_i$ are uniquely determined. So, we've established a bijection: for each $m_1 = 25, 26, \ldots, 50,$ we want to count the number of ordered tuples $(a_1, a_2, \ldots, a_9)$ of nonnegative integers such that

$a_1 + a_2 + \ldots + a_9 \le 50-m_1.$

Let this be $f(m_1)$ ; then, we want to find the value of

$\displaystyle\sum_{m_1=25}^{50} f(m_1).$

Now, if $a_1 + a_2 + \ldots + a_9 \le 50 - m_1,$ then there must be some nonnegative integer $x$ so that

$a_1 + a_2 + \ldots + a_9 + x = 50 - m_1.$

Interpret $x$ as the difference between $50-m_1$ and $a_1 + \ldots + a_9.$ Then, we see that each possible choice of $(a_1, \ldots, a_9)$ corresponds to a choice of $(a_1, \ldots, a_9, x),$ and vice versa (another bijection!) By the balls-and-urns (AKA stars-and-bars) approach, there are $\dbinom{59-m_1}{9}$ ways to pick $a_1, a_2, \ldots, a_9, x,$ so there are $\dbinom{59-m_1}{9}$ ways to pick $a_1, a_2, \ldots, a_9,$ under our original condition. Thus,

$f(m_1) = \dbinom{59-m_1}{9}.$

Then we have

$\begin{aligned} \displaystyle\sum_{m_1=25}^{50} f(m_1) &= \displaystyle\sum_{m_1=25}^{50} \dbinom{59-m_1}{9} \\ &= \dbinom{34}{9} + \dbinom{33}{9} + \ldots + \dbinom{9}{9}, \end{aligned}$ which equals $\dbinom{35}{10} = 183579396$ by the Hockey Stick Identity. The answer is $\boxed{396}.$

First, we notice that if we choose 10 passing scores (from 25 to 50), there is exactly 1 unique ordering of those scores such that the scores are in a non-decreasing order. So, we just have to find a way to choose all the unique combinations of 10 scores from 25 to 50 to solve this problem, since each combination translates to exactly 1 permutation. Thus, we have turned our permutation problem into a combination problem.

Next, we realize that this is simply a stars and bars problem--we have 10 exams, and we need to assign each one a score, or place it into a "score bin," if you will. This is as if we had 10 balls which we had to place into 26 boxes, except the balls are exams and the boxes are scores (from 25 to 50). So then, using stars and bars, the number of stars is 10, and the number of bars is 25, which means that the solution is simply ${25+10 \choose 10}$ , or ${35 \choose 10}$ , the last 3 digits of which are 396 .

Using Stars and Bars (combinatorics) concept, we get N = C(10+26-1, 26-1) = C(35, 25) = C(35, 10) = 183579396

The last 3 digits of N i.e. N (mod 1000) = 396.

Let Peter's marks in the $i$ th year be $m_i$ . Clearly, from the condition specified, $25 \leq m_1 \leq m_2 \leq \cdots \leq m_{10} \leq 50$ . The number of solutions to this equation is ${10 + 25 \choose 10} = \boxed{183579396}$ . You may refer Jatin's post for details of the number of solutions part.

Applying the combinations formula (with repetition), or i dont know how's called in english... Anyways, that is:

${n + r - 1 \choose r}$

So in this case we substitute $n = 26$ (since from $25$ to $50$ there are $50 - 25 + 1 = 26$ numbers), and $r = 10$ .

$\frac {35!}{10!\cdot 25!}$

The big question is: "how am I going to compute this without calculator?" Well, let's try to simplify in some way...

$\frac {26 \cdot 27 \cdot \ldots \cdot 34 \cdot 35}{10!}$

$= \frac {13\cdot 9\cdot 7\cdot 29\cdot 6\cdot 31\cdot 32\cdot 33\cdot 34\cdot 5}{6\cdot 8\cdot 9\cdot 10}$

$= 2^{2}\cdot 3\cdot 7\cdot 11\cdot 13\cdot 17\cdot 29\cdot 31$

That I turned to the form

$12012\cdot 15283$

You know.. With a lot of patience we get

$183579396$

Thus, the last $3$ digits of $N$ are

$\boxed {396}$ .

number of compositions of 36 into 11 parts