Geometrical Permutations - Practice Questions & MCQ

Rule for Geometrical Permutations -

There are n points in the plane such that no three of them are on the same straight line except m of them which are on same straight line, then the number of straight lines formed by joining them is/are ${ }^n c_2-{ }^m c_2+1$ and number triangles is/are ${ }^n c_3-{ }^m c_3$.

Let's say there is a round table with 6 chairs all identical and 6 persons have to sit. For the first person there is only one choice to make as all chairs are identical, so wherever he may sit doesn't matter. Now when $1^{\text {st }}$ person has sit, $2^{\text {nd }}$ person with respect to $1^{\text {st }}$ have five choices to sit, directly opposite or left or $2^{\text {nd }}$ from left or right or $2^{\text {nd }}$ from right to $1^{\text {st }}$ person, in the same ways $3^{\text {rd }}$ person will have 4 choices, $4^{\text {th }}$ person will have 3 choice, $5^{\text {th }}$ person will have 2 choices, and last person 1 choice, so in that way total $5 \times 4 \times 3 \times 2 \times 1=(6-1)$ ! permutations are possible.

We can generalize this result as an object that can be arranged along with a circular table in (n-1)! Ways

Example: In how many ways 7 people can be arranged along a circular table having 7 identical chairs.

Solution: using the above concept, it can be done in (7-1)! = $6!\mid$
Necklace, Garland Type Questions
If clockwise and anticlockwise permutations are same in a circular permutation, (as in case of garlands or necklace formation), the number of permutations becomes $(1 / 2)(n-1)$ !

Since as in the case of necklace and garland if we flip the bead or garland then anticlockwise arrangements become clockwise but they are identical because the objects are identical, hence two arrangements are reduced to one causing the total number of permutations to be halved.

Example: Find the ways in which 10 different beads can be arranged to form a necklace?
Solution: Using circular permutations the total number of permutations $=(10-1)$ !
Now since clockwise and anticlockwise arrangements give the same permutation so the total number of permutations becomes $(1 / 2)(9!)$

If clockwise and anticlockwise permutations are same in a circular permutation, (as in case of garlands or necklace formation), the number of permutations becomes (1/2)(n-1)!.

Since as in the case of necklace and garland if we flip the bead or garland then anticlockwise arrangements become clockwise but they are identical because the objects are identical, hence two arrangements are reduced to one causing the total number of permutations to be halved.

Example: Find the ways in which 10 different beads can be arranged to form a necklace?
Solution: Using circular permutations the total number of permutations $=(10-1)$ !
Now since clockwise and anticlockwise arrangements give the same permutation so the total number of permutations becomes (1/2)(9!)