118929
There are 10 points in a plane, out of these 6 are collinear. If \(N\) is the total number of triangles formed by joining these points, then \(\mathrm{N}\) \(=\)
1 120
2 850
3 100
4 150
Explanation:
C Given, 10 points in a plane, out of which 6 are collinear. \(\mathrm{N}\) is the total number of triangle formed by joining these points. Case I:- When all 10 points are collinear, then number of triangles formed, \({ }^{10} \mathrm{C}_3=\frac{10 !}{3 ! 7 !} =\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 7 !}\) \(=120\) Case II:- When 6 points are collinear then number of triangle are formed, \(\quad{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2}=20\) \(\therefore \text { Required number of triangle } =120-20\) \(=100\)
AP EAMCET-05.07.2022
Permutation and Combination
118930
A polygon has 54 diagonals. The number of sides of this polygon is.......
1 12
2 15
3 16
4 9
Explanation:
A The number of diagonals of a polygon \(=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-3)}{2}=54\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}=108\) \(\Rightarrow \quad \mathrm{n}^2-3 \mathrm{n}-108=0\) \(\mathrm{n}^2-12 \mathrm{n}+9 \mathrm{n}-108=0\) \(\mathrm{n}(\mathrm{n}-12)+9(\mathrm{n}-12)=0\) \((\mathrm{n}-12) \quad \mathrm{n}+9)=0\) \(\Rightarrow \quad \mathrm{n}-12=0\) \(\Rightarrow \quad \mathrm{n}=12\) \(\text { If } \mathrm{n}+9=0\) \(\Rightarrow \quad \mathrm{n}=-9\) \(\text { The number of sides cannot be negative therefore we }\) \(\text { neglect } \mathrm{n}=-9\) \(\text { Hence, } \mathrm{n}=12\)
AP EAMCET-18.09.2020
Permutation and Combination
118931
If \({ }^{\mathrm{n}} \mathbf{C}_7={ }^{\mathrm{n}} \mathbf{C}_6\), then \({ }^{\mathrm{n}} \mathbf{C}_2=\)
118929
There are 10 points in a plane, out of these 6 are collinear. If \(N\) is the total number of triangles formed by joining these points, then \(\mathrm{N}\) \(=\)
1 120
2 850
3 100
4 150
Explanation:
C Given, 10 points in a plane, out of which 6 are collinear. \(\mathrm{N}\) is the total number of triangle formed by joining these points. Case I:- When all 10 points are collinear, then number of triangles formed, \({ }^{10} \mathrm{C}_3=\frac{10 !}{3 ! 7 !} =\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 7 !}\) \(=120\) Case II:- When 6 points are collinear then number of triangle are formed, \(\quad{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2}=20\) \(\therefore \text { Required number of triangle } =120-20\) \(=100\)
AP EAMCET-05.07.2022
Permutation and Combination
118930
A polygon has 54 diagonals. The number of sides of this polygon is.......
1 12
2 15
3 16
4 9
Explanation:
A The number of diagonals of a polygon \(=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-3)}{2}=54\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}=108\) \(\Rightarrow \quad \mathrm{n}^2-3 \mathrm{n}-108=0\) \(\mathrm{n}^2-12 \mathrm{n}+9 \mathrm{n}-108=0\) \(\mathrm{n}(\mathrm{n}-12)+9(\mathrm{n}-12)=0\) \((\mathrm{n}-12) \quad \mathrm{n}+9)=0\) \(\Rightarrow \quad \mathrm{n}-12=0\) \(\Rightarrow \quad \mathrm{n}=12\) \(\text { If } \mathrm{n}+9=0\) \(\Rightarrow \quad \mathrm{n}=-9\) \(\text { The number of sides cannot be negative therefore we }\) \(\text { neglect } \mathrm{n}=-9\) \(\text { Hence, } \mathrm{n}=12\)
AP EAMCET-18.09.2020
Permutation and Combination
118931
If \({ }^{\mathrm{n}} \mathbf{C}_7={ }^{\mathrm{n}} \mathbf{C}_6\), then \({ }^{\mathrm{n}} \mathbf{C}_2=\)
118929
There are 10 points in a plane, out of these 6 are collinear. If \(N\) is the total number of triangles formed by joining these points, then \(\mathrm{N}\) \(=\)
1 120
2 850
3 100
4 150
Explanation:
C Given, 10 points in a plane, out of which 6 are collinear. \(\mathrm{N}\) is the total number of triangle formed by joining these points. Case I:- When all 10 points are collinear, then number of triangles formed, \({ }^{10} \mathrm{C}_3=\frac{10 !}{3 ! 7 !} =\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 7 !}\) \(=120\) Case II:- When 6 points are collinear then number of triangle are formed, \(\quad{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2}=20\) \(\therefore \text { Required number of triangle } =120-20\) \(=100\)
AP EAMCET-05.07.2022
Permutation and Combination
118930
A polygon has 54 diagonals. The number of sides of this polygon is.......
1 12
2 15
3 16
4 9
Explanation:
A The number of diagonals of a polygon \(=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-3)}{2}=54\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}=108\) \(\Rightarrow \quad \mathrm{n}^2-3 \mathrm{n}-108=0\) \(\mathrm{n}^2-12 \mathrm{n}+9 \mathrm{n}-108=0\) \(\mathrm{n}(\mathrm{n}-12)+9(\mathrm{n}-12)=0\) \((\mathrm{n}-12) \quad \mathrm{n}+9)=0\) \(\Rightarrow \quad \mathrm{n}-12=0\) \(\Rightarrow \quad \mathrm{n}=12\) \(\text { If } \mathrm{n}+9=0\) \(\Rightarrow \quad \mathrm{n}=-9\) \(\text { The number of sides cannot be negative therefore we }\) \(\text { neglect } \mathrm{n}=-9\) \(\text { Hence, } \mathrm{n}=12\)
AP EAMCET-18.09.2020
Permutation and Combination
118931
If \({ }^{\mathrm{n}} \mathbf{C}_7={ }^{\mathrm{n}} \mathbf{C}_6\), then \({ }^{\mathrm{n}} \mathbf{C}_2=\)
118929
There are 10 points in a plane, out of these 6 are collinear. If \(N\) is the total number of triangles formed by joining these points, then \(\mathrm{N}\) \(=\)
1 120
2 850
3 100
4 150
Explanation:
C Given, 10 points in a plane, out of which 6 are collinear. \(\mathrm{N}\) is the total number of triangle formed by joining these points. Case I:- When all 10 points are collinear, then number of triangles formed, \({ }^{10} \mathrm{C}_3=\frac{10 !}{3 ! 7 !} =\frac{10 \times 9 \times 8 \times 7 !}{3 \times 2 \times 7 !}\) \(=120\) Case II:- When 6 points are collinear then number of triangle are formed, \(\quad{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2}=20\) \(\therefore \text { Required number of triangle } =120-20\) \(=100\)
AP EAMCET-05.07.2022
Permutation and Combination
118930
A polygon has 54 diagonals. The number of sides of this polygon is.......
1 12
2 15
3 16
4 9
Explanation:
A The number of diagonals of a polygon \(=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\text { So, } \quad \frac{\mathrm{n}(\mathrm{n}-3)}{2}=54\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}=108\) \(\Rightarrow \quad \mathrm{n}^2-3 \mathrm{n}-108=0\) \(\mathrm{n}^2-12 \mathrm{n}+9 \mathrm{n}-108=0\) \(\mathrm{n}(\mathrm{n}-12)+9(\mathrm{n}-12)=0\) \((\mathrm{n}-12) \quad \mathrm{n}+9)=0\) \(\Rightarrow \quad \mathrm{n}-12=0\) \(\Rightarrow \quad \mathrm{n}=12\) \(\text { If } \mathrm{n}+9=0\) \(\Rightarrow \quad \mathrm{n}=-9\) \(\text { The number of sides cannot be negative therefore we }\) \(\text { neglect } \mathrm{n}=-9\) \(\text { Hence, } \mathrm{n}=12\)
AP EAMCET-18.09.2020
Permutation and Combination
118931
If \({ }^{\mathrm{n}} \mathbf{C}_7={ }^{\mathrm{n}} \mathbf{C}_6\), then \({ }^{\mathrm{n}} \mathbf{C}_2=\)
