118949
If \(p\) and \(q\) are positive integers such that \({ }^{(p+q)} P_2\) \(=42\) and \({ }^{(p-q)} P_2=20\), the values of \(p\) and \(q\) are respectively
1 5,2
2 4,3
3 7,2
4 6,1
5 7,5
Explanation:
D Given that, \({ }^{(p+q)} P_2=42\) and \({ }^{(p-q)} P_2=20\) We know that, From equation, \(\mathrm{p}+\mathrm{q}=7\) \(\mathrm{p}+6=7\) \(\mathrm{p}=7-6\) \(\mathrm{q}=6, \mathrm{p}=1\) \(\therefore \quad \mathrm{q}=6, \mathrm{p}=1\)
Kerala CEE-2021
Permutation and Combination
118924
If \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) then the minimum value of \(n\) is
1 5
2 6
3 7
4 8
Explanation:
D Given, \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) We know that- \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\) \(\Rightarrow{ }^n C_4>{ }^n C_3\) \(\Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !}\) \(\Rightarrow \frac{1}{4}>\frac{1}{n-3}\) \(\Rightarrow n-3>4\) \(\Rightarrow n>7\)So, the minimum value of \(\mathrm{n}\) is 8 .
AP EAMCET-2011
Permutation and Combination
118925
A polygon has 35 diagonals, then the number of its sides is
1 8
2 9
3 10
4 11
Explanation:
C Given that, polygon has \(35 \text { diagonals }\) \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \quad(\because \mathrm{n} \text { number of sides })\) \(35=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\mathrm{n}(\mathrm{n}-3)=70\) \(\mathrm{n}^2-3 \mathrm{n}-70=0\) \(\mathrm{n}^2-10 \mathrm{n}+7 \mathrm{n}-70=0\) \(\mathrm{n}(\mathrm{n}-10)+7(\mathrm{n}-10)=0\) \((\mathrm{n}-10)(\mathrm{n}+7)=0\) \(\mathrm{n}=10,-7\) \(\text { Since, the number of side cannot be negative therefore } \mathrm{n}\) \(=10\)
118949
If \(p\) and \(q\) are positive integers such that \({ }^{(p+q)} P_2\) \(=42\) and \({ }^{(p-q)} P_2=20\), the values of \(p\) and \(q\) are respectively
1 5,2
2 4,3
3 7,2
4 6,1
5 7,5
Explanation:
D Given that, \({ }^{(p+q)} P_2=42\) and \({ }^{(p-q)} P_2=20\) We know that, From equation, \(\mathrm{p}+\mathrm{q}=7\) \(\mathrm{p}+6=7\) \(\mathrm{p}=7-6\) \(\mathrm{q}=6, \mathrm{p}=1\) \(\therefore \quad \mathrm{q}=6, \mathrm{p}=1\)
Kerala CEE-2021
Permutation and Combination
118924
If \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) then the minimum value of \(n\) is
1 5
2 6
3 7
4 8
Explanation:
D Given, \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) We know that- \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\) \(\Rightarrow{ }^n C_4>{ }^n C_3\) \(\Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !}\) \(\Rightarrow \frac{1}{4}>\frac{1}{n-3}\) \(\Rightarrow n-3>4\) \(\Rightarrow n>7\)So, the minimum value of \(\mathrm{n}\) is 8 .
AP EAMCET-2011
Permutation and Combination
118925
A polygon has 35 diagonals, then the number of its sides is
1 8
2 9
3 10
4 11
Explanation:
C Given that, polygon has \(35 \text { diagonals }\) \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \quad(\because \mathrm{n} \text { number of sides })\) \(35=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\mathrm{n}(\mathrm{n}-3)=70\) \(\mathrm{n}^2-3 \mathrm{n}-70=0\) \(\mathrm{n}^2-10 \mathrm{n}+7 \mathrm{n}-70=0\) \(\mathrm{n}(\mathrm{n}-10)+7(\mathrm{n}-10)=0\) \((\mathrm{n}-10)(\mathrm{n}+7)=0\) \(\mathrm{n}=10,-7\) \(\text { Since, the number of side cannot be negative therefore } \mathrm{n}\) \(=10\)
118949
If \(p\) and \(q\) are positive integers such that \({ }^{(p+q)} P_2\) \(=42\) and \({ }^{(p-q)} P_2=20\), the values of \(p\) and \(q\) are respectively
1 5,2
2 4,3
3 7,2
4 6,1
5 7,5
Explanation:
D Given that, \({ }^{(p+q)} P_2=42\) and \({ }^{(p-q)} P_2=20\) We know that, From equation, \(\mathrm{p}+\mathrm{q}=7\) \(\mathrm{p}+6=7\) \(\mathrm{p}=7-6\) \(\mathrm{q}=6, \mathrm{p}=1\) \(\therefore \quad \mathrm{q}=6, \mathrm{p}=1\)
Kerala CEE-2021
Permutation and Combination
118924
If \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) then the minimum value of \(n\) is
1 5
2 6
3 7
4 8
Explanation:
D Given, \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) We know that- \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\) \(\Rightarrow{ }^n C_4>{ }^n C_3\) \(\Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !}\) \(\Rightarrow \frac{1}{4}>\frac{1}{n-3}\) \(\Rightarrow n-3>4\) \(\Rightarrow n>7\)So, the minimum value of \(\mathrm{n}\) is 8 .
AP EAMCET-2011
Permutation and Combination
118925
A polygon has 35 diagonals, then the number of its sides is
1 8
2 9
3 10
4 11
Explanation:
C Given that, polygon has \(35 \text { diagonals }\) \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \quad(\because \mathrm{n} \text { number of sides })\) \(35=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\mathrm{n}(\mathrm{n}-3)=70\) \(\mathrm{n}^2-3 \mathrm{n}-70=0\) \(\mathrm{n}^2-10 \mathrm{n}+7 \mathrm{n}-70=0\) \(\mathrm{n}(\mathrm{n}-10)+7(\mathrm{n}-10)=0\) \((\mathrm{n}-10)(\mathrm{n}+7)=0\) \(\mathrm{n}=10,-7\) \(\text { Since, the number of side cannot be negative therefore } \mathrm{n}\) \(=10\)
118949
If \(p\) and \(q\) are positive integers such that \({ }^{(p+q)} P_2\) \(=42\) and \({ }^{(p-q)} P_2=20\), the values of \(p\) and \(q\) are respectively
1 5,2
2 4,3
3 7,2
4 6,1
5 7,5
Explanation:
D Given that, \({ }^{(p+q)} P_2=42\) and \({ }^{(p-q)} P_2=20\) We know that, From equation, \(\mathrm{p}+\mathrm{q}=7\) \(\mathrm{p}+6=7\) \(\mathrm{p}=7-6\) \(\mathrm{q}=6, \mathrm{p}=1\) \(\therefore \quad \mathrm{q}=6, \mathrm{p}=1\)
Kerala CEE-2021
Permutation and Combination
118924
If \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) then the minimum value of \(n\) is
1 5
2 6
3 7
4 8
Explanation:
D Given, \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) We know that- \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\) \(\Rightarrow{ }^n C_4>{ }^n C_3\) \(\Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !}\) \(\Rightarrow \frac{1}{4}>\frac{1}{n-3}\) \(\Rightarrow n-3>4\) \(\Rightarrow n>7\)So, the minimum value of \(\mathrm{n}\) is 8 .
AP EAMCET-2011
Permutation and Combination
118925
A polygon has 35 diagonals, then the number of its sides is
1 8
2 9
3 10
4 11
Explanation:
C Given that, polygon has \(35 \text { diagonals }\) \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \quad(\because \mathrm{n} \text { number of sides })\) \(35=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\mathrm{n}(\mathrm{n}-3)=70\) \(\mathrm{n}^2-3 \mathrm{n}-70=0\) \(\mathrm{n}^2-10 \mathrm{n}+7 \mathrm{n}-70=0\) \(\mathrm{n}(\mathrm{n}-10)+7(\mathrm{n}-10)=0\) \((\mathrm{n}-10)(\mathrm{n}+7)=0\) \(\mathrm{n}=10,-7\) \(\text { Since, the number of side cannot be negative therefore } \mathrm{n}\) \(=10\)
118949
If \(p\) and \(q\) are positive integers such that \({ }^{(p+q)} P_2\) \(=42\) and \({ }^{(p-q)} P_2=20\), the values of \(p\) and \(q\) are respectively
1 5,2
2 4,3
3 7,2
4 6,1
5 7,5
Explanation:
D Given that, \({ }^{(p+q)} P_2=42\) and \({ }^{(p-q)} P_2=20\) We know that, From equation, \(\mathrm{p}+\mathrm{q}=7\) \(\mathrm{p}+6=7\) \(\mathrm{p}=7-6\) \(\mathrm{q}=6, \mathrm{p}=1\) \(\therefore \quad \mathrm{q}=6, \mathrm{p}=1\)
Kerala CEE-2021
Permutation and Combination
118924
If \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) then the minimum value of \(n\) is
1 5
2 6
3 7
4 8
Explanation:
D Given, \({ }^{(\mathrm{n}-1)} \mathrm{C}_3+{ }^{(\mathrm{n}-1)} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3\) We know that- \({ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\) \(\Rightarrow{ }^n C_4>{ }^n C_3\) \(\Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !}\) \(\Rightarrow \frac{1}{4}>\frac{1}{n-3}\) \(\Rightarrow n-3>4\) \(\Rightarrow n>7\)So, the minimum value of \(\mathrm{n}\) is 8 .
AP EAMCET-2011
Permutation and Combination
118925
A polygon has 35 diagonals, then the number of its sides is
1 8
2 9
3 10
4 11
Explanation:
C Given that, polygon has \(35 \text { diagonals }\) \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2} \quad(\because \mathrm{n} \text { number of sides })\) \(35=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\mathrm{n}(\mathrm{n}-3)=70\) \(\mathrm{n}^2-3 \mathrm{n}-70=0\) \(\mathrm{n}^2-10 \mathrm{n}+7 \mathrm{n}-70=0\) \(\mathrm{n}(\mathrm{n}-10)+7(\mathrm{n}-10)=0\) \((\mathrm{n}-10)(\mathrm{n}+7)=0\) \(\mathrm{n}=10,-7\) \(\text { Since, the number of side cannot be negative therefore } \mathrm{n}\) \(=10\)