A We have, \(\frac{{ }^n P_{r-1}}{a}=\frac{{ }^n P_r}{b}=\frac{{ }^n P_{r+1}}{c}\) \(\frac{1}{\mathrm{a}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{1}{\mathrm{~b}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{1}{\mathrm{c}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}-1) !}\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{n}-\mathrm{r}+1\) \(\frac{\mathrm{c}}{\mathrm{b}}=\mathrm{n}-\mathrm{r}\) \(\text { From equation (i) and (ii), we get- }\) \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}+1 \Rightarrow \mathrm{b}^2=\mathrm{a}(\mathrm{b}+\mathrm{c})\) From equation (i) and (ii), we get-
COMEDK-2019
Permutation and Combination
119249
If \(\frac{n !}{5 !(n-1) !}\) and \(\frac{n !}{7 !(n-3) !}\) are in the ratio 21 : 1 , then find the value of \(n\),
1 1
2 2
3 3
4 4
Explanation:
C We have, \(\frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !}: \frac{\mathrm{n} !}{7 !(\mathrm{n}-3) !}=21: 1\) \(\Rightarrow \frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !} \times \frac{7 !(\mathrm{n}-3) !}{\mathrm{n} !}=\frac{21}{1}\) \(\Rightarrow \frac{7 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1) !}=\frac{21}{1}\) \(\frac{7 \times 6 \times 5 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}=\frac{21}{1}\) \(\Rightarrow \frac{42}{\mathrm{n}^2-3 \mathrm{n}+2}=\frac{21}{1}\) \(\Rightarrow \mathrm{n}(\mathrm{n}-3)=0 \Rightarrow \mathrm{n}=0 \text { or } \mathrm{n}=3\) \(\text { But when } \mathrm{n}=0 \text {, then }(\mathrm{n}-1) ! \text { and }(\mathrm{n}-3) \text { !are not defined. }\) \(\text { Thus, } \mathrm{n}=3\) But when \(\mathrm{n}=0\), then ( \(\mathrm{n}-1\) )! and ( \(\mathrm{n}-3)\) !are not defined. Thus, \(\mathrm{n}=3\)
COMEDK-2019
Permutation and Combination
119250
The number of integral solutions of \(x+y+z=\) 0, with \(x \geq-5, y \geq-5, z \geq-5\), is
1 135
2 136
3 455
4 105
Explanation:
B We have, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=0\) Where, \(x \geq-5, y \geq-5, z \geq-5\) Now put :- \(\mathrm{x}=\mathrm{t}_1-5, \mathrm{y}=\mathrm{t}_2-5\) and \(\mathrm{z}=\mathrm{t}_3-5\) \(\therefore \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=15\) where \(\mathrm{t}_1, \mathrm{t}_2, \mathrm{t}_3 \geq 0\) Required number of solutions - \(={ }^{15+3-1} \mathrm{C}_{3-1}\) \(={ }^{17} \mathrm{C}_2\) \(=\frac{17 !}{2 ! 15 !}=\frac{17 \times 16 \times 15 !}{2 \times 1 \times 15 !}=136\)
SRM JEEE-2008
Permutation and Combination
119251
The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is
1 24
2 52
3 48
4 16
Explanation:
D Required no. of triangles \(=\) total no. of triangles - no. of triangles having one side common - no. of triangles having two sides common. \(={ }^8 \mathrm{C}_3-8 \times 4-8\) \(=\frac{8 !}{3 ! 5 !}-32-8=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}-32-8\) \(=56-32-8=56-40=16\)
A We have, \(\frac{{ }^n P_{r-1}}{a}=\frac{{ }^n P_r}{b}=\frac{{ }^n P_{r+1}}{c}\) \(\frac{1}{\mathrm{a}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{1}{\mathrm{~b}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{1}{\mathrm{c}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}-1) !}\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{n}-\mathrm{r}+1\) \(\frac{\mathrm{c}}{\mathrm{b}}=\mathrm{n}-\mathrm{r}\) \(\text { From equation (i) and (ii), we get- }\) \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}+1 \Rightarrow \mathrm{b}^2=\mathrm{a}(\mathrm{b}+\mathrm{c})\) From equation (i) and (ii), we get-
COMEDK-2019
Permutation and Combination
119249
If \(\frac{n !}{5 !(n-1) !}\) and \(\frac{n !}{7 !(n-3) !}\) are in the ratio 21 : 1 , then find the value of \(n\),
1 1
2 2
3 3
4 4
Explanation:
C We have, \(\frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !}: \frac{\mathrm{n} !}{7 !(\mathrm{n}-3) !}=21: 1\) \(\Rightarrow \frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !} \times \frac{7 !(\mathrm{n}-3) !}{\mathrm{n} !}=\frac{21}{1}\) \(\Rightarrow \frac{7 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1) !}=\frac{21}{1}\) \(\frac{7 \times 6 \times 5 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}=\frac{21}{1}\) \(\Rightarrow \frac{42}{\mathrm{n}^2-3 \mathrm{n}+2}=\frac{21}{1}\) \(\Rightarrow \mathrm{n}(\mathrm{n}-3)=0 \Rightarrow \mathrm{n}=0 \text { or } \mathrm{n}=3\) \(\text { But when } \mathrm{n}=0 \text {, then }(\mathrm{n}-1) ! \text { and }(\mathrm{n}-3) \text { !are not defined. }\) \(\text { Thus, } \mathrm{n}=3\) But when \(\mathrm{n}=0\), then ( \(\mathrm{n}-1\) )! and ( \(\mathrm{n}-3)\) !are not defined. Thus, \(\mathrm{n}=3\)
COMEDK-2019
Permutation and Combination
119250
The number of integral solutions of \(x+y+z=\) 0, with \(x \geq-5, y \geq-5, z \geq-5\), is
1 135
2 136
3 455
4 105
Explanation:
B We have, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=0\) Where, \(x \geq-5, y \geq-5, z \geq-5\) Now put :- \(\mathrm{x}=\mathrm{t}_1-5, \mathrm{y}=\mathrm{t}_2-5\) and \(\mathrm{z}=\mathrm{t}_3-5\) \(\therefore \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=15\) where \(\mathrm{t}_1, \mathrm{t}_2, \mathrm{t}_3 \geq 0\) Required number of solutions - \(={ }^{15+3-1} \mathrm{C}_{3-1}\) \(={ }^{17} \mathrm{C}_2\) \(=\frac{17 !}{2 ! 15 !}=\frac{17 \times 16 \times 15 !}{2 \times 1 \times 15 !}=136\)
SRM JEEE-2008
Permutation and Combination
119251
The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is
1 24
2 52
3 48
4 16
Explanation:
D Required no. of triangles \(=\) total no. of triangles - no. of triangles having one side common - no. of triangles having two sides common. \(={ }^8 \mathrm{C}_3-8 \times 4-8\) \(=\frac{8 !}{3 ! 5 !}-32-8=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}-32-8\) \(=56-32-8=56-40=16\)
A We have, \(\frac{{ }^n P_{r-1}}{a}=\frac{{ }^n P_r}{b}=\frac{{ }^n P_{r+1}}{c}\) \(\frac{1}{\mathrm{a}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{1}{\mathrm{~b}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{1}{\mathrm{c}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}-1) !}\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{n}-\mathrm{r}+1\) \(\frac{\mathrm{c}}{\mathrm{b}}=\mathrm{n}-\mathrm{r}\) \(\text { From equation (i) and (ii), we get- }\) \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}+1 \Rightarrow \mathrm{b}^2=\mathrm{a}(\mathrm{b}+\mathrm{c})\) From equation (i) and (ii), we get-
COMEDK-2019
Permutation and Combination
119249
If \(\frac{n !}{5 !(n-1) !}\) and \(\frac{n !}{7 !(n-3) !}\) are in the ratio 21 : 1 , then find the value of \(n\),
1 1
2 2
3 3
4 4
Explanation:
C We have, \(\frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !}: \frac{\mathrm{n} !}{7 !(\mathrm{n}-3) !}=21: 1\) \(\Rightarrow \frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !} \times \frac{7 !(\mathrm{n}-3) !}{\mathrm{n} !}=\frac{21}{1}\) \(\Rightarrow \frac{7 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1) !}=\frac{21}{1}\) \(\frac{7 \times 6 \times 5 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}=\frac{21}{1}\) \(\Rightarrow \frac{42}{\mathrm{n}^2-3 \mathrm{n}+2}=\frac{21}{1}\) \(\Rightarrow \mathrm{n}(\mathrm{n}-3)=0 \Rightarrow \mathrm{n}=0 \text { or } \mathrm{n}=3\) \(\text { But when } \mathrm{n}=0 \text {, then }(\mathrm{n}-1) ! \text { and }(\mathrm{n}-3) \text { !are not defined. }\) \(\text { Thus, } \mathrm{n}=3\) But when \(\mathrm{n}=0\), then ( \(\mathrm{n}-1\) )! and ( \(\mathrm{n}-3)\) !are not defined. Thus, \(\mathrm{n}=3\)
COMEDK-2019
Permutation and Combination
119250
The number of integral solutions of \(x+y+z=\) 0, with \(x \geq-5, y \geq-5, z \geq-5\), is
1 135
2 136
3 455
4 105
Explanation:
B We have, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=0\) Where, \(x \geq-5, y \geq-5, z \geq-5\) Now put :- \(\mathrm{x}=\mathrm{t}_1-5, \mathrm{y}=\mathrm{t}_2-5\) and \(\mathrm{z}=\mathrm{t}_3-5\) \(\therefore \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=15\) where \(\mathrm{t}_1, \mathrm{t}_2, \mathrm{t}_3 \geq 0\) Required number of solutions - \(={ }^{15+3-1} \mathrm{C}_{3-1}\) \(={ }^{17} \mathrm{C}_2\) \(=\frac{17 !}{2 ! 15 !}=\frac{17 \times 16 \times 15 !}{2 \times 1 \times 15 !}=136\)
SRM JEEE-2008
Permutation and Combination
119251
The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is
1 24
2 52
3 48
4 16
Explanation:
D Required no. of triangles \(=\) total no. of triangles - no. of triangles having one side common - no. of triangles having two sides common. \(={ }^8 \mathrm{C}_3-8 \times 4-8\) \(=\frac{8 !}{3 ! 5 !}-32-8=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}-32-8\) \(=56-32-8=56-40=16\)
A We have, \(\frac{{ }^n P_{r-1}}{a}=\frac{{ }^n P_r}{b}=\frac{{ }^n P_{r+1}}{c}\) \(\frac{1}{\mathrm{a}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}+1) !}=\frac{1}{\mathrm{~b}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}=\frac{1}{\mathrm{c}} \frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}-1) !}\) \(\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\mathrm{n}-\mathrm{r}+1\) \(\frac{\mathrm{c}}{\mathrm{b}}=\mathrm{n}-\mathrm{r}\) \(\text { From equation (i) and (ii), we get- }\) \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}+1 \Rightarrow \mathrm{b}^2=\mathrm{a}(\mathrm{b}+\mathrm{c})\) From equation (i) and (ii), we get-
COMEDK-2019
Permutation and Combination
119249
If \(\frac{n !}{5 !(n-1) !}\) and \(\frac{n !}{7 !(n-3) !}\) are in the ratio 21 : 1 , then find the value of \(n\),
1 1
2 2
3 3
4 4
Explanation:
C We have, \(\frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !}: \frac{\mathrm{n} !}{7 !(\mathrm{n}-3) !}=21: 1\) \(\Rightarrow \frac{\mathrm{n} !}{5 !(\mathrm{n}-1) !} \times \frac{7 !(\mathrm{n}-3) !}{\mathrm{n} !}=\frac{21}{1}\) \(\Rightarrow \frac{7 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1) !}=\frac{21}{1}\) \(\frac{7 \times 6 \times 5 !(\mathrm{n}-3) !}{5 !(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) !}=\frac{21}{1}\) \(\Rightarrow \frac{42}{\mathrm{n}^2-3 \mathrm{n}+2}=\frac{21}{1}\) \(\Rightarrow \mathrm{n}(\mathrm{n}-3)=0 \Rightarrow \mathrm{n}=0 \text { or } \mathrm{n}=3\) \(\text { But when } \mathrm{n}=0 \text {, then }(\mathrm{n}-1) ! \text { and }(\mathrm{n}-3) \text { !are not defined. }\) \(\text { Thus, } \mathrm{n}=3\) But when \(\mathrm{n}=0\), then ( \(\mathrm{n}-1\) )! and ( \(\mathrm{n}-3)\) !are not defined. Thus, \(\mathrm{n}=3\)
COMEDK-2019
Permutation and Combination
119250
The number of integral solutions of \(x+y+z=\) 0, with \(x \geq-5, y \geq-5, z \geq-5\), is
1 135
2 136
3 455
4 105
Explanation:
B We have, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=0\) Where, \(x \geq-5, y \geq-5, z \geq-5\) Now put :- \(\mathrm{x}=\mathrm{t}_1-5, \mathrm{y}=\mathrm{t}_2-5\) and \(\mathrm{z}=\mathrm{t}_3-5\) \(\therefore \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=15\) where \(\mathrm{t}_1, \mathrm{t}_2, \mathrm{t}_3 \geq 0\) Required number of solutions - \(={ }^{15+3-1} \mathrm{C}_{3-1}\) \(={ }^{17} \mathrm{C}_2\) \(=\frac{17 !}{2 ! 15 !}=\frac{17 \times 16 \times 15 !}{2 \times 1 \times 15 !}=136\)
SRM JEEE-2008
Permutation and Combination
119251
The number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the sides of the octagon is
1 24
2 52
3 48
4 16
Explanation:
D Required no. of triangles \(=\) total no. of triangles - no. of triangles having one side common - no. of triangles having two sides common. \(={ }^8 \mathrm{C}_3-8 \times 4-8\) \(=\frac{8 !}{3 ! 5 !}-32-8=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}-32-8\) \(=56-32-8=56-40=16\)
