119282
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(T_{n+1}-T_n=10\), then the value of \(n\) is :
1 5
2 3
3 7
4 4
Explanation:
A 3 vertices needs for a triangle out of \(n\) vertices should be chosen \(={ }^n C_3\) According to question \({ }^{n+1} C_3-{ }^n C_3=10\) \(\frac{(n+1) !}{3 !(n-2) !}-\frac{n !}{3 !(n-3) !}=10\) \({ }^n C_2+{ }^n C_3-{ }^n C_3=10\) \({ }^n C_2=10\) \(\frac{n !}{2 !(n-2) !}=10\) \(\mathrm{n}(\mathrm{n}-1)=20\) \(\mathrm{n}^2-\mathrm{n}-20=0\) \(\mathrm{n}^2-5 \mathrm{n}+4 \mathrm{n}-20=0\) \((\mathrm{n}-5)(\mathrm{n}+4)=0\) \(\mathrm{n}-5=0\) \(\mathrm{n}=5\)
AP EAMCET-06.07.2022
Permutation and Combination
119283
If a polygon of \(n\) sides has 560 diagonals, then \(\mathbf{n -}\)
1 35
2 36
3 37
4 38
Explanation:
A Here number of diagonals =560 \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 560=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 1120=\mathrm{n}^2-3 \mathrm{n}\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}^2-35 \mathrm{n}+32 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}(\mathrm{n}-35)+32(\mathrm{n}-35)=0\) \(\Rightarrow(\mathrm{n}-35)(\mathrm{n}+32)=0\) \(\Rightarrow \mathrm{n}=35 \quad(\because \mathrm{n} \neq-32)\)
AP EAMCET-04.07.2021
Permutation and Combination
119284
The total number of permutations of \(n\) different things taken not more than \(r\) at a time, when each thing may be repeated any number of times is
C Total number of things \(=\mathrm{n}\) The number of ways of taken 1 at a time \(=n\) Similarly 2 times \(=\mathrm{n}^2\) 3 times \(=n^3\) \(\mathrm{r}\) times \(=\mathrm{n}^{\mathrm{r}}\) \(\therefore\) Total number of ways taking \(1,2, \ldots \ldots . . \mathrm{r}\) of a time \(=\mathrm{n}+\mathrm{n}^2+\mathrm{n}^3+\ldots \ldots \ldots \ldots \mathrm{n}^{\mathrm{r}}\) \(=\mathrm{n}\left(1+\mathrm{n}+\mathrm{n}^2+\ldots \ldots \ldots . \mathrm{n}^{\mathrm{r}-1}\right)\) \(=\mathrm{n}\left(\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots . .1\right)\) Multiply by \(\frac{(\mathrm{n}-1)}{(\mathrm{n}-1)}\) \(=\mathrm{n}\left(\frac{\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots \ldots+1}{\mathrm{n}-1}\right)(\mathrm{n}-1)=\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{r}}-1\right)}{(\mathrm{n}-1)}\)
AP EAMCET-04.07.2021
Permutation and Combination
119285
Let \(m\) be a natural number such that \(20000\lt \) \(m\lt 60000\) and let \(k\) be the sum of all the digits in \(m\). then the number of numbers \(m\) for which \(k\) is even, is
1 19909
2 19989
3 18999
4 19999
Explanation:
D Natural numbers are \(1,2,3,4, \ldots . . .9\) Let as consider 10 successive five digit numbers \(\begin{array}{ll}\mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 0 \\ \mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 1 \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 2 \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots & \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 9\end{array}\) Where \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \ldots \ldots . . \) are some digits 5 different values of which the sum of all digits is even. \(\therefore \text { Value of } \mathrm{k} \text { is }=4 \times 10^3 \times 5-1\) \(=20 \times 10^3-1\) \(=20000-1\) \(=19999\)
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Permutation and Combination
119282
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(T_{n+1}-T_n=10\), then the value of \(n\) is :
1 5
2 3
3 7
4 4
Explanation:
A 3 vertices needs for a triangle out of \(n\) vertices should be chosen \(={ }^n C_3\) According to question \({ }^{n+1} C_3-{ }^n C_3=10\) \(\frac{(n+1) !}{3 !(n-2) !}-\frac{n !}{3 !(n-3) !}=10\) \({ }^n C_2+{ }^n C_3-{ }^n C_3=10\) \({ }^n C_2=10\) \(\frac{n !}{2 !(n-2) !}=10\) \(\mathrm{n}(\mathrm{n}-1)=20\) \(\mathrm{n}^2-\mathrm{n}-20=0\) \(\mathrm{n}^2-5 \mathrm{n}+4 \mathrm{n}-20=0\) \((\mathrm{n}-5)(\mathrm{n}+4)=0\) \(\mathrm{n}-5=0\) \(\mathrm{n}=5\)
AP EAMCET-06.07.2022
Permutation and Combination
119283
If a polygon of \(n\) sides has 560 diagonals, then \(\mathbf{n -}\)
1 35
2 36
3 37
4 38
Explanation:
A Here number of diagonals =560 \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 560=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 1120=\mathrm{n}^2-3 \mathrm{n}\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}^2-35 \mathrm{n}+32 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}(\mathrm{n}-35)+32(\mathrm{n}-35)=0\) \(\Rightarrow(\mathrm{n}-35)(\mathrm{n}+32)=0\) \(\Rightarrow \mathrm{n}=35 \quad(\because \mathrm{n} \neq-32)\)
AP EAMCET-04.07.2021
Permutation and Combination
119284
The total number of permutations of \(n\) different things taken not more than \(r\) at a time, when each thing may be repeated any number of times is
C Total number of things \(=\mathrm{n}\) The number of ways of taken 1 at a time \(=n\) Similarly 2 times \(=\mathrm{n}^2\) 3 times \(=n^3\) \(\mathrm{r}\) times \(=\mathrm{n}^{\mathrm{r}}\) \(\therefore\) Total number of ways taking \(1,2, \ldots \ldots . . \mathrm{r}\) of a time \(=\mathrm{n}+\mathrm{n}^2+\mathrm{n}^3+\ldots \ldots \ldots \ldots \mathrm{n}^{\mathrm{r}}\) \(=\mathrm{n}\left(1+\mathrm{n}+\mathrm{n}^2+\ldots \ldots \ldots . \mathrm{n}^{\mathrm{r}-1}\right)\) \(=\mathrm{n}\left(\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots . .1\right)\) Multiply by \(\frac{(\mathrm{n}-1)}{(\mathrm{n}-1)}\) \(=\mathrm{n}\left(\frac{\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots \ldots+1}{\mathrm{n}-1}\right)(\mathrm{n}-1)=\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{r}}-1\right)}{(\mathrm{n}-1)}\)
AP EAMCET-04.07.2021
Permutation and Combination
119285
Let \(m\) be a natural number such that \(20000\lt \) \(m\lt 60000\) and let \(k\) be the sum of all the digits in \(m\). then the number of numbers \(m\) for which \(k\) is even, is
1 19909
2 19989
3 18999
4 19999
Explanation:
D Natural numbers are \(1,2,3,4, \ldots . . .9\) Let as consider 10 successive five digit numbers \(\begin{array}{ll}\mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 0 \\ \mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 1 \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 2 \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots & \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 9\end{array}\) Where \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \ldots \ldots . . \) are some digits 5 different values of which the sum of all digits is even. \(\therefore \text { Value of } \mathrm{k} \text { is }=4 \times 10^3 \times 5-1\) \(=20 \times 10^3-1\) \(=20000-1\) \(=19999\)
119282
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(T_{n+1}-T_n=10\), then the value of \(n\) is :
1 5
2 3
3 7
4 4
Explanation:
A 3 vertices needs for a triangle out of \(n\) vertices should be chosen \(={ }^n C_3\) According to question \({ }^{n+1} C_3-{ }^n C_3=10\) \(\frac{(n+1) !}{3 !(n-2) !}-\frac{n !}{3 !(n-3) !}=10\) \({ }^n C_2+{ }^n C_3-{ }^n C_3=10\) \({ }^n C_2=10\) \(\frac{n !}{2 !(n-2) !}=10\) \(\mathrm{n}(\mathrm{n}-1)=20\) \(\mathrm{n}^2-\mathrm{n}-20=0\) \(\mathrm{n}^2-5 \mathrm{n}+4 \mathrm{n}-20=0\) \((\mathrm{n}-5)(\mathrm{n}+4)=0\) \(\mathrm{n}-5=0\) \(\mathrm{n}=5\)
AP EAMCET-06.07.2022
Permutation and Combination
119283
If a polygon of \(n\) sides has 560 diagonals, then \(\mathbf{n -}\)
1 35
2 36
3 37
4 38
Explanation:
A Here number of diagonals =560 \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 560=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 1120=\mathrm{n}^2-3 \mathrm{n}\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}^2-35 \mathrm{n}+32 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}(\mathrm{n}-35)+32(\mathrm{n}-35)=0\) \(\Rightarrow(\mathrm{n}-35)(\mathrm{n}+32)=0\) \(\Rightarrow \mathrm{n}=35 \quad(\because \mathrm{n} \neq-32)\)
AP EAMCET-04.07.2021
Permutation and Combination
119284
The total number of permutations of \(n\) different things taken not more than \(r\) at a time, when each thing may be repeated any number of times is
C Total number of things \(=\mathrm{n}\) The number of ways of taken 1 at a time \(=n\) Similarly 2 times \(=\mathrm{n}^2\) 3 times \(=n^3\) \(\mathrm{r}\) times \(=\mathrm{n}^{\mathrm{r}}\) \(\therefore\) Total number of ways taking \(1,2, \ldots \ldots . . \mathrm{r}\) of a time \(=\mathrm{n}+\mathrm{n}^2+\mathrm{n}^3+\ldots \ldots \ldots \ldots \mathrm{n}^{\mathrm{r}}\) \(=\mathrm{n}\left(1+\mathrm{n}+\mathrm{n}^2+\ldots \ldots \ldots . \mathrm{n}^{\mathrm{r}-1}\right)\) \(=\mathrm{n}\left(\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots . .1\right)\) Multiply by \(\frac{(\mathrm{n}-1)}{(\mathrm{n}-1)}\) \(=\mathrm{n}\left(\frac{\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots \ldots+1}{\mathrm{n}-1}\right)(\mathrm{n}-1)=\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{r}}-1\right)}{(\mathrm{n}-1)}\)
AP EAMCET-04.07.2021
Permutation and Combination
119285
Let \(m\) be a natural number such that \(20000\lt \) \(m\lt 60000\) and let \(k\) be the sum of all the digits in \(m\). then the number of numbers \(m\) for which \(k\) is even, is
1 19909
2 19989
3 18999
4 19999
Explanation:
D Natural numbers are \(1,2,3,4, \ldots . . .9\) Let as consider 10 successive five digit numbers \(\begin{array}{ll}\mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 0 \\ \mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 1 \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 2 \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots & \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 9\end{array}\) Where \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \ldots \ldots . . \) are some digits 5 different values of which the sum of all digits is even. \(\therefore \text { Value of } \mathrm{k} \text { is }=4 \times 10^3 \times 5-1\) \(=20 \times 10^3-1\) \(=20000-1\) \(=19999\)
119282
Let \(T_n\) be the number of all possible triangles formed by joining vertices of an \(n\)-sided regular polygon. If \(T_{n+1}-T_n=10\), then the value of \(n\) is :
1 5
2 3
3 7
4 4
Explanation:
A 3 vertices needs for a triangle out of \(n\) vertices should be chosen \(={ }^n C_3\) According to question \({ }^{n+1} C_3-{ }^n C_3=10\) \(\frac{(n+1) !}{3 !(n-2) !}-\frac{n !}{3 !(n-3) !}=10\) \({ }^n C_2+{ }^n C_3-{ }^n C_3=10\) \({ }^n C_2=10\) \(\frac{n !}{2 !(n-2) !}=10\) \(\mathrm{n}(\mathrm{n}-1)=20\) \(\mathrm{n}^2-\mathrm{n}-20=0\) \(\mathrm{n}^2-5 \mathrm{n}+4 \mathrm{n}-20=0\) \((\mathrm{n}-5)(\mathrm{n}+4)=0\) \(\mathrm{n}-5=0\) \(\mathrm{n}=5\)
AP EAMCET-06.07.2022
Permutation and Combination
119283
If a polygon of \(n\) sides has 560 diagonals, then \(\mathbf{n -}\)
1 35
2 36
3 37
4 38
Explanation:
A Here number of diagonals =560 \(\text { Number of diagonals }=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 560=\frac{\mathrm{n}(\mathrm{n}-3)}{2}\) \(\Rightarrow \quad 1120=\mathrm{n}^2-3 \mathrm{n}\) \(\Rightarrow \mathrm{n}^2-3 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}^2-35 \mathrm{n}+32 \mathrm{n}-1120=0\) \(\Rightarrow \mathrm{n}(\mathrm{n}-35)+32(\mathrm{n}-35)=0\) \(\Rightarrow(\mathrm{n}-35)(\mathrm{n}+32)=0\) \(\Rightarrow \mathrm{n}=35 \quad(\because \mathrm{n} \neq-32)\)
AP EAMCET-04.07.2021
Permutation and Combination
119284
The total number of permutations of \(n\) different things taken not more than \(r\) at a time, when each thing may be repeated any number of times is
C Total number of things \(=\mathrm{n}\) The number of ways of taken 1 at a time \(=n\) Similarly 2 times \(=\mathrm{n}^2\) 3 times \(=n^3\) \(\mathrm{r}\) times \(=\mathrm{n}^{\mathrm{r}}\) \(\therefore\) Total number of ways taking \(1,2, \ldots \ldots . . \mathrm{r}\) of a time \(=\mathrm{n}+\mathrm{n}^2+\mathrm{n}^3+\ldots \ldots \ldots \ldots \mathrm{n}^{\mathrm{r}}\) \(=\mathrm{n}\left(1+\mathrm{n}+\mathrm{n}^2+\ldots \ldots \ldots . \mathrm{n}^{\mathrm{r}-1}\right)\) \(=\mathrm{n}\left(\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots . .1\right)\) Multiply by \(\frac{(\mathrm{n}-1)}{(\mathrm{n}-1)}\) \(=\mathrm{n}\left(\frac{\mathrm{n}^{\mathrm{r}-1}+\mathrm{n}^{\mathrm{r}-2}+\ldots \ldots \ldots+1}{\mathrm{n}-1}\right)(\mathrm{n}-1)=\frac{\mathrm{n}\left(\mathrm{n}^{\mathrm{r}}-1\right)}{(\mathrm{n}-1)}\)
AP EAMCET-04.07.2021
Permutation and Combination
119285
Let \(m\) be a natural number such that \(20000\lt \) \(m\lt 60000\) and let \(k\) be the sum of all the digits in \(m\). then the number of numbers \(m\) for which \(k\) is even, is
1 19909
2 19989
3 18999
4 19999
Explanation:
D Natural numbers are \(1,2,3,4, \ldots . . .9\) Let as consider 10 successive five digit numbers \(\begin{array}{ll}\mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 0 \\ \mathrm{x}_1, \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 1 \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 2 \\ \ldots \ldots \ldots \ldots \ldots \ldots \ldots & \\ \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 & 9\end{array}\) Where \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \ldots \ldots . . \) are some digits 5 different values of which the sum of all digits is even. \(\therefore \text { Value of } \mathrm{k} \text { is }=4 \times 10^3 \times 5-1\) \(=20 \times 10^3-1\) \(=20000-1\) \(=19999\)