# CRE 10 - Geometry based questions | Modern Math - Permutation & Combination

**Answer the next 2 questions based on the information given:**

Out of 15 points on a plane 5 are collinear, except which no other set of 3 points is collinear. By joining these points:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

How many straight lines can be formed?

Answer: 96

**Explanation** :

If we select any two non-collinear points we will get a different line except when we select the 5 collinear points, we will get the same line again and again.

Total number of ways of choosing 2 points = ^{15}C_{2} = 105.

For collinear set of points, number of selecting 2 points = ^{5}C_{2} = 10.

Now each of these 10 lines is same hence, we will subtract these 10 lines and add 1 line.

∴ Required answer = 105 – 10 + 1 = 96.

Hence, 96.

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

How many triangles can be formed?

Answer: 445

**Explanation** :

A triangle can formed by choosing any 3 points expect all 3 are collinear when we won’t get any triangle.

∴ Total number of triangles = ^{15}C_{3} – ^{5}C_{3} = 455 – 10 = 445.

Hence, 445.

Workspace:

**Answer the next 4 questions based on the information given:**

In the given 8 × 8 grid:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Find the number of squares.

Answer: 204

**Explanation** :

Number of squares in a m × n grid = 1^{2} + 2^{2} + 3^{2} + … + n^{2}. (where m ≥ n)

∴ Number of squares in this 8 × 8 grid = 1^{2} + 2^{2} + 3^{2} + … + 8^{2} = 204.

Hence, 204.

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Find the number of rectangles.

Answer: 1296

**Explanation** :

Number of rectangles in a m × n grid = ^{m+1}C_{2} × ^{n+1}C_{2}.

∴ Number of rectangles in a 8 × 8 grid = ^{9}C_{2} × ^{9}C_{2} = 36 × 36 = 1296.

Hence, 1296.

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Find the number of ways in which a person can travel from A to C, if the person can only travel on a grid line either towards the right or upwards.

- A.
16!

- B.
$\frac{16!}{8!\times 8!}$

- C.
^{16}C_{8} - D.
None of these

Answer: Option B

**Explanation** :

To go from A to C, a person has to take 8 steps toward right and 8 steps upwards.

One of the ways is: R R R R R R R R U U U U U U U U

R → 1 step towards right

U → 1 step upward.

Total number of ways of going from A to C is same as total ways of arranging these 8 identical Rs and 8 identical Us.

∴ Total number of ways = 16!/(8!×8!).

Hence, option (b).

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Find the number of ways in which a person can travel from A to B via C, if the person can only travel on a grid line either towards the right or upwards.

- A.
$\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$

- B.
$\frac{16!}{8!\times 8!}$

- C.
8! × 8!

- D.
None of these

Answer: Option A

**Explanation** :

Consider the solution to the previous question.

**A → B: **To go from A to B, we need to take 4 steps towards right and 4 steps upwards i.e., total number of ways = $\frac{8!}{4!\times 4!}$

**B → C:** To go from B to C, we need to take 4 steps towards right and 4 steps upwards i.e., total number of ways = $\frac{8!}{4!\times 4!}$

**A → B → C:** To go from A to C via B, total number of ways = $\frac{8!}{4!\times 4!}\times \frac{8!}{4!\times 4!}$ = $\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$.

Hence, option (a).

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Find the number of ways in which a person can travel from A to B without going via C, if the person can only travel on a grid line either towards the right or upwards.

- A.
$\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$

- B.
16! - 8!

- C.
$\frac{16!}{8!\times 8!}-\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$

- D.
None of these

Answer: Option C

**Explanation** :

Total number of ways of going from A to C = $\frac{16!}{8!\times 8!}$.

Total number of ways of going from A to C via B = $\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$.

Total number of ways of going from A to C without going through B = $\frac{16!}{8!\times 8!}-\frac{{\left(8!\right)}^{2}}{{\left(4!\right)}^{4}}$.

Hence, option (c).

Workspace:

**Answer the next 2 questions based on the information given:**

There are 10 straight lines on a plane. Find the

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Minimum number of bounded and unbounded regions on the plane.

- A.
0 and 0

- B.
0 and 11

- C.
10 and 11

- D.
None of these

Answer: Option B

**Explanation** :

For n lines in a plane:

**Minimum Bounded **regions will be 0 when all the lines are parallel to each other.

**Minimum Unbounded** regions will be (n + 1) when all the lines are parallel to each other.

Hence, option (b).

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Maximum number of total regions on the plane.

Answer: 56

**Explanation** :

Maximum number of regions can be obtained when all lines intersect each other at distinct points i.e., $\frac{n\left(n+1\right)}{2}$+1.

∴ Required answer = $\frac{10\left(10+1\right)}{2}$ + 1 = 56.

Hence, 56.

Workspace:

**CRE 10 - Geometry based questions | Modern Math - Permutation & Combination**

Minimum number of bounded and unbounded regions on the plane.

- A.
10 and 36

- B.
28 and 28

- C.
36 and 20

- D.
None of these

Answer: Option C

**Explanation** :

For n lines in a plane:

Maximum Unbounded regions is 2n = 20, when not all lines are parallel.

Maximum Bounded region = n(n+1)/2 + 1 - 2n = 56 – 20 = 36

Hence, option (c).

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**