What is so special about $111$ ?

Probability Level 5

Consider a regular $111$ sided polygon. We will get a decagon by joining any $10$ points of this $111$ sided polygon.

In this polygon, how many decagon can be formed such that they do not have any side common with the $111$ sided polygon?

If the answer contains $m$ digits and sum of all digits of the answer is $n$ , find $m + n$ .

$\text{Details and Assumptions :}$

$\bullet$ Sum of digits means sum of every individual digits. For example, sum of digits in number $122453$ is $1+2+2+4+5+3 = 17$

$\bullet$ You may use computational tools

The answer is 56.

1 solution

Christopher Boo
Mar 29, 2015

This problem is actually similar to the Fastidious Librarian , since the simplest wording of the problem is

We need to choose 10 from 111 vertices such that no two of them are consecutive.

An alternative interpretation of the problem is

We need to insert 10 vertices into 101 vertices such that no two of them are consecutive.

We see the 101 vertices as lined-up balls and 10 to-be-inserted vertices as bars. Note that there cannot be bars at the leftmost and rightmost position at the same time.

We first calculate the number of ways to insert 10 bars into 102 gaps $102\choose3$ without concerning if the leftmost and rightmost position is filled, then minus the number of ways to insert 8 bars into 100 gaps $100\choose8$ when the leftmost and rightmost position is occupied.

Hence,

${102\choose10} - {100\choose8} = 21114773073240$

There's nothing special about $111$ , it's just the art of counting.

Let me quote:

The generalization of the problem is left as an exercise for the reader.

Christopher Boo - 6 years, 2 months ago

Genaralized result for r-sided polygon to be fitted in n-sided polygon is,

$\displaystyle\frac{n}{r} {n-r-1 \choose r-1}$

Akshay Bodhare - 6 years, 2 months ago

What is so special about 111 111 1 1 1 ?

The answer is 56.

1 solution

0 pending reports

What is so special about $111$ ?