119256
There are 10 points in a plane, out of these 10 points 6 are collinear. The number of triangles formed by joining these points is
1 120
2 150
3 100
4 180
Explanation:
C We have, 6 points are collinear Number of triangle formed if all 10 point are non- collinear is \({ }^{10} \mathrm{C}_3\). No. of triangles formed \({ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=120-20\) \(=100\)
SRM JEEE-2010
Permutation and Combination
119258
A polygon has 44 diagonals. The number of sides is
1 0
2 11
3 12
4 9
Explanation:
B Here, Let the number of sides be \(\mathrm{n}\) \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=44\) \(\Rightarrow \mathrm{n}^2-\mathrm{n}-2 \mathrm{n}=88\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-88=0\) \(\Rightarrow (\mathrm{n}-11)(\mathrm{n}+8)=0\) \(\Rightarrow \mathrm{n}=11 \quad(\because \mathrm{n} \neq 8)\) \(\therefore \text { Number of sides }=11\) \(\therefore \quad\) Number of sides \(=11\)
SRM JEEE-2013
Permutation and Combination
119259
930 Deepawali greeting cards are exchanged amongst the students of a class. If every student sends a card to every other student, then what is the number of students in the class?
1 31
2 29
3 43
4 24 [SRJMJEEE-2012]
Explanation:
A :We have, Number of deepawali greeting cards \(=930\) Let the number of students in the class be \(\mathrm{n}\). \(\therefore\) Number of cards send by \(\mathrm{n}\) students \(=\mathrm{n}(\mathrm{n}-1)\) According to question, \(\mathrm{n}(\mathrm{n}-1)=930\) \(\mathrm{n}^2-\mathrm{n}-930=0\) \((\mathrm{n}-31)(\mathrm{n}+30)=0\) \(\mathrm{n}=31\) \(\therefore \text { No. of students in class }=31 \quad(\because \mathrm{n} \neq-30)\)
Permutation and Combination
119260
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is
1 4820
2 1410
3 2830
4 none of these
Explanation:
D Consider the arrangement by taking two girls together as one and hence the 7 girls can now be arranged in \(7 !=5040\) \(\therefore\) Total ways in which two girls sit together \(=5040 \times 2=10080\) The required no. of ways \(=\) the number of ways in which 8 girls can sit - the no. of ways in which two sisters are together \(=8 !-(2) 7 !=40320-2 \times 5040\) \(=40320-10080=30240\)
SRM JEEE-2014
Permutation and Combination
119261
If the total number of \(m\) elements subsets of the set \(A=\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \ldots . \mathbf{a}_{\mathbf{n}}\right\}\) is \(\lambda\) times the number of 3 elements subsets containing \(a_4\), then \(n\) is
1 \((\mathrm{m}-1) \lambda\)
2 \(\mathrm{m} \lambda\)
3 \((\mathrm{m}+1) \lambda\)
4 0
Explanation:
B Here From set of \(n\) element selecting a subset of \(\mathrm{m}\) element \(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}}\) Total number of subsets of \(A\) each containing the element \(\mathrm{a}_4={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) Thus, according to question, \({ }^n \mathrm{C}_{\mathrm{m}}=\lambda .{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \frac{\mathrm{n}}{\mathrm{m}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}=\lambda \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \lambda=\frac{\mathrm{n}}{\mathrm{m}}\) \(\Rightarrow \mathrm{n}=\mathrm{m} \lambda .\)
119256
There are 10 points in a plane, out of these 10 points 6 are collinear. The number of triangles formed by joining these points is
1 120
2 150
3 100
4 180
Explanation:
C We have, 6 points are collinear Number of triangle formed if all 10 point are non- collinear is \({ }^{10} \mathrm{C}_3\). No. of triangles formed \({ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=120-20\) \(=100\)
SRM JEEE-2010
Permutation and Combination
119258
A polygon has 44 diagonals. The number of sides is
1 0
2 11
3 12
4 9
Explanation:
B Here, Let the number of sides be \(\mathrm{n}\) \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=44\) \(\Rightarrow \mathrm{n}^2-\mathrm{n}-2 \mathrm{n}=88\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-88=0\) \(\Rightarrow (\mathrm{n}-11)(\mathrm{n}+8)=0\) \(\Rightarrow \mathrm{n}=11 \quad(\because \mathrm{n} \neq 8)\) \(\therefore \text { Number of sides }=11\) \(\therefore \quad\) Number of sides \(=11\)
SRM JEEE-2013
Permutation and Combination
119259
930 Deepawali greeting cards are exchanged amongst the students of a class. If every student sends a card to every other student, then what is the number of students in the class?
1 31
2 29
3 43
4 24 [SRJMJEEE-2012]
Explanation:
A :We have, Number of deepawali greeting cards \(=930\) Let the number of students in the class be \(\mathrm{n}\). \(\therefore\) Number of cards send by \(\mathrm{n}\) students \(=\mathrm{n}(\mathrm{n}-1)\) According to question, \(\mathrm{n}(\mathrm{n}-1)=930\) \(\mathrm{n}^2-\mathrm{n}-930=0\) \((\mathrm{n}-31)(\mathrm{n}+30)=0\) \(\mathrm{n}=31\) \(\therefore \text { No. of students in class }=31 \quad(\because \mathrm{n} \neq-30)\)
Permutation and Combination
119260
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is
1 4820
2 1410
3 2830
4 none of these
Explanation:
D Consider the arrangement by taking two girls together as one and hence the 7 girls can now be arranged in \(7 !=5040\) \(\therefore\) Total ways in which two girls sit together \(=5040 \times 2=10080\) The required no. of ways \(=\) the number of ways in which 8 girls can sit - the no. of ways in which two sisters are together \(=8 !-(2) 7 !=40320-2 \times 5040\) \(=40320-10080=30240\)
SRM JEEE-2014
Permutation and Combination
119261
If the total number of \(m\) elements subsets of the set \(A=\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \ldots . \mathbf{a}_{\mathbf{n}}\right\}\) is \(\lambda\) times the number of 3 elements subsets containing \(a_4\), then \(n\) is
1 \((\mathrm{m}-1) \lambda\)
2 \(\mathrm{m} \lambda\)
3 \((\mathrm{m}+1) \lambda\)
4 0
Explanation:
B Here From set of \(n\) element selecting a subset of \(\mathrm{m}\) element \(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}}\) Total number of subsets of \(A\) each containing the element \(\mathrm{a}_4={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) Thus, according to question, \({ }^n \mathrm{C}_{\mathrm{m}}=\lambda .{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \frac{\mathrm{n}}{\mathrm{m}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}=\lambda \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \lambda=\frac{\mathrm{n}}{\mathrm{m}}\) \(\Rightarrow \mathrm{n}=\mathrm{m} \lambda .\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Permutation and Combination
119256
There are 10 points in a plane, out of these 10 points 6 are collinear. The number of triangles formed by joining these points is
1 120
2 150
3 100
4 180
Explanation:
C We have, 6 points are collinear Number of triangle formed if all 10 point are non- collinear is \({ }^{10} \mathrm{C}_3\). No. of triangles formed \({ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=120-20\) \(=100\)
SRM JEEE-2010
Permutation and Combination
119258
A polygon has 44 diagonals. The number of sides is
1 0
2 11
3 12
4 9
Explanation:
B Here, Let the number of sides be \(\mathrm{n}\) \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=44\) \(\Rightarrow \mathrm{n}^2-\mathrm{n}-2 \mathrm{n}=88\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-88=0\) \(\Rightarrow (\mathrm{n}-11)(\mathrm{n}+8)=0\) \(\Rightarrow \mathrm{n}=11 \quad(\because \mathrm{n} \neq 8)\) \(\therefore \text { Number of sides }=11\) \(\therefore \quad\) Number of sides \(=11\)
SRM JEEE-2013
Permutation and Combination
119259
930 Deepawali greeting cards are exchanged amongst the students of a class. If every student sends a card to every other student, then what is the number of students in the class?
1 31
2 29
3 43
4 24 [SRJMJEEE-2012]
Explanation:
A :We have, Number of deepawali greeting cards \(=930\) Let the number of students in the class be \(\mathrm{n}\). \(\therefore\) Number of cards send by \(\mathrm{n}\) students \(=\mathrm{n}(\mathrm{n}-1)\) According to question, \(\mathrm{n}(\mathrm{n}-1)=930\) \(\mathrm{n}^2-\mathrm{n}-930=0\) \((\mathrm{n}-31)(\mathrm{n}+30)=0\) \(\mathrm{n}=31\) \(\therefore \text { No. of students in class }=31 \quad(\because \mathrm{n} \neq-30)\)
Permutation and Combination
119260
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is
1 4820
2 1410
3 2830
4 none of these
Explanation:
D Consider the arrangement by taking two girls together as one and hence the 7 girls can now be arranged in \(7 !=5040\) \(\therefore\) Total ways in which two girls sit together \(=5040 \times 2=10080\) The required no. of ways \(=\) the number of ways in which 8 girls can sit - the no. of ways in which two sisters are together \(=8 !-(2) 7 !=40320-2 \times 5040\) \(=40320-10080=30240\)
SRM JEEE-2014
Permutation and Combination
119261
If the total number of \(m\) elements subsets of the set \(A=\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \ldots . \mathbf{a}_{\mathbf{n}}\right\}\) is \(\lambda\) times the number of 3 elements subsets containing \(a_4\), then \(n\) is
1 \((\mathrm{m}-1) \lambda\)
2 \(\mathrm{m} \lambda\)
3 \((\mathrm{m}+1) \lambda\)
4 0
Explanation:
B Here From set of \(n\) element selecting a subset of \(\mathrm{m}\) element \(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}}\) Total number of subsets of \(A\) each containing the element \(\mathrm{a}_4={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) Thus, according to question, \({ }^n \mathrm{C}_{\mathrm{m}}=\lambda .{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \frac{\mathrm{n}}{\mathrm{m}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}=\lambda \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \lambda=\frac{\mathrm{n}}{\mathrm{m}}\) \(\Rightarrow \mathrm{n}=\mathrm{m} \lambda .\)
119256
There are 10 points in a plane, out of these 10 points 6 are collinear. The number of triangles formed by joining these points is
1 120
2 150
3 100
4 180
Explanation:
C We have, 6 points are collinear Number of triangle formed if all 10 point are non- collinear is \({ }^{10} \mathrm{C}_3\). No. of triangles formed \({ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=120-20\) \(=100\)
SRM JEEE-2010
Permutation and Combination
119258
A polygon has 44 diagonals. The number of sides is
1 0
2 11
3 12
4 9
Explanation:
B Here, Let the number of sides be \(\mathrm{n}\) \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=44\) \(\Rightarrow \mathrm{n}^2-\mathrm{n}-2 \mathrm{n}=88\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-88=0\) \(\Rightarrow (\mathrm{n}-11)(\mathrm{n}+8)=0\) \(\Rightarrow \mathrm{n}=11 \quad(\because \mathrm{n} \neq 8)\) \(\therefore \text { Number of sides }=11\) \(\therefore \quad\) Number of sides \(=11\)
SRM JEEE-2013
Permutation and Combination
119259
930 Deepawali greeting cards are exchanged amongst the students of a class. If every student sends a card to every other student, then what is the number of students in the class?
1 31
2 29
3 43
4 24 [SRJMJEEE-2012]
Explanation:
A :We have, Number of deepawali greeting cards \(=930\) Let the number of students in the class be \(\mathrm{n}\). \(\therefore\) Number of cards send by \(\mathrm{n}\) students \(=\mathrm{n}(\mathrm{n}-1)\) According to question, \(\mathrm{n}(\mathrm{n}-1)=930\) \(\mathrm{n}^2-\mathrm{n}-930=0\) \((\mathrm{n}-31)(\mathrm{n}+30)=0\) \(\mathrm{n}=31\) \(\therefore \text { No. of students in class }=31 \quad(\because \mathrm{n} \neq-30)\)
Permutation and Combination
119260
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is
1 4820
2 1410
3 2830
4 none of these
Explanation:
D Consider the arrangement by taking two girls together as one and hence the 7 girls can now be arranged in \(7 !=5040\) \(\therefore\) Total ways in which two girls sit together \(=5040 \times 2=10080\) The required no. of ways \(=\) the number of ways in which 8 girls can sit - the no. of ways in which two sisters are together \(=8 !-(2) 7 !=40320-2 \times 5040\) \(=40320-10080=30240\)
SRM JEEE-2014
Permutation and Combination
119261
If the total number of \(m\) elements subsets of the set \(A=\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \ldots . \mathbf{a}_{\mathbf{n}}\right\}\) is \(\lambda\) times the number of 3 elements subsets containing \(a_4\), then \(n\) is
1 \((\mathrm{m}-1) \lambda\)
2 \(\mathrm{m} \lambda\)
3 \((\mathrm{m}+1) \lambda\)
4 0
Explanation:
B Here From set of \(n\) element selecting a subset of \(\mathrm{m}\) element \(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}}\) Total number of subsets of \(A\) each containing the element \(\mathrm{a}_4={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) Thus, according to question, \({ }^n \mathrm{C}_{\mathrm{m}}=\lambda .{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \frac{\mathrm{n}}{\mathrm{m}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}=\lambda \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \lambda=\frac{\mathrm{n}}{\mathrm{m}}\) \(\Rightarrow \mathrm{n}=\mathrm{m} \lambda .\)
119256
There are 10 points in a plane, out of these 10 points 6 are collinear. The number of triangles formed by joining these points is
1 120
2 150
3 100
4 180
Explanation:
C We have, 6 points are collinear Number of triangle formed if all 10 point are non- collinear is \({ }^{10} \mathrm{C}_3\). No. of triangles formed \({ }^{10} \mathrm{C}_3-{ }^6 \mathrm{C}_3\) \(=120-20\) \(=100\)
SRM JEEE-2010
Permutation and Combination
119258
A polygon has 44 diagonals. The number of sides is
1 0
2 11
3 12
4 9
Explanation:
B Here, Let the number of sides be \(\mathrm{n}\) \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2-\mathrm{n}=44\) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}-\mathrm{n}=44\) \(\Rightarrow \mathrm{n}^2-\mathrm{n}-2 \mathrm{n}=88\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-88=0\) \(\Rightarrow (\mathrm{n}-11)(\mathrm{n}+8)=0\) \(\Rightarrow \mathrm{n}=11 \quad(\because \mathrm{n} \neq 8)\) \(\therefore \text { Number of sides }=11\) \(\therefore \quad\) Number of sides \(=11\)
SRM JEEE-2013
Permutation and Combination
119259
930 Deepawali greeting cards are exchanged amongst the students of a class. If every student sends a card to every other student, then what is the number of students in the class?
1 31
2 29
3 43
4 24 [SRJMJEEE-2012]
Explanation:
A :We have, Number of deepawali greeting cards \(=930\) Let the number of students in the class be \(\mathrm{n}\). \(\therefore\) Number of cards send by \(\mathrm{n}\) students \(=\mathrm{n}(\mathrm{n}-1)\) According to question, \(\mathrm{n}(\mathrm{n}-1)=930\) \(\mathrm{n}^2-\mathrm{n}-930=0\) \((\mathrm{n}-31)(\mathrm{n}+30)=0\) \(\mathrm{n}=31\) \(\therefore \text { No. of students in class }=31 \quad(\because \mathrm{n} \neq-30)\)
Permutation and Combination
119260
In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is
1 4820
2 1410
3 2830
4 none of these
Explanation:
D Consider the arrangement by taking two girls together as one and hence the 7 girls can now be arranged in \(7 !=5040\) \(\therefore\) Total ways in which two girls sit together \(=5040 \times 2=10080\) The required no. of ways \(=\) the number of ways in which 8 girls can sit - the no. of ways in which two sisters are together \(=8 !-(2) 7 !=40320-2 \times 5040\) \(=40320-10080=30240\)
SRM JEEE-2014
Permutation and Combination
119261
If the total number of \(m\) elements subsets of the set \(A=\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3, \ldots . \mathbf{a}_{\mathbf{n}}\right\}\) is \(\lambda\) times the number of 3 elements subsets containing \(a_4\), then \(n\) is
1 \((\mathrm{m}-1) \lambda\)
2 \(\mathrm{m} \lambda\)
3 \((\mathrm{m}+1) \lambda\)
4 0
Explanation:
B Here From set of \(n\) element selecting a subset of \(\mathrm{m}\) element \(={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{m}}\) Total number of subsets of \(A\) each containing the element \(\mathrm{a}_4={ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) Thus, according to question, \({ }^n \mathrm{C}_{\mathrm{m}}=\lambda .{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \frac{\mathrm{n}}{\mathrm{m}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}=\lambda \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{m}-1}\) \(\Rightarrow \lambda=\frac{\mathrm{n}}{\mathrm{m}}\) \(\Rightarrow \mathrm{n}=\mathrm{m} \lambda .\)