119267
\(P(n)=n(n+1)(n+5), n \in N\) is a multiple of
1 9
2 5
3 3
4 7
Explanation:
C : Here, \(p(n)=n(n+1)(n+5)\) \(P(1)=1(2)(6)=12=4 \times 3\) \(P(2)=2(3)(7)=42=14 \times 3\) \(P(3)=3(4)(8)=96=32 \times 3\)3 is common multiple in all \(n \in \mathrm{N}\).
JCECE-2019
Permutation and Combination
119268
The value of \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is
1 \(\frac{2^{50}}{51}\)
2 \(\frac{2^{50}-1}{51}\)
3 \(\frac{2^{50}-1}{50}\)
4 None of these
Explanation:
A Given that, \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots . .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is \(=\left(\frac{50 !}{\frac{50 !}{1}}+\frac{50 !}{\frac{2 ! 48 !}{3}}+\ldots \ldots \frac{50 !}{\frac{50 ! 0 !}{51}}\right)\) \(=\left(1+\frac{50 \times 49}{3 \times 2 !}+\ldots \ldots .+\frac{1}{51}\right)\) Multiply the above expression by \(\frac{51}{51}\) \(=\frac{1}{51}\left(51+\frac{51 \times 50 \times 99}{3 !}+\ldots . .\right)\) \(=\frac{1}{51}\left({ }^{51} \mathrm{C}_1+{ }^{51} \mathrm{C}_3+\ldots .\right)\) As, we know that sum of odd coefficient \(=2^{\mathrm{n}-1}\) \(=\frac{1}{51} 2^{51-1}=\frac{2^{50}}{51}\)
BCECE-2014
Permutation and Combination
119269
If \({ }^n C_{r-1}=10,{ }^n C_r=45\), and \({ }^n C_{r+1}=120\), then \(r\) equals to
1 1
2 2
3 3
4 4
Explanation:
B : We have, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=10\) \({ }^n \mathrm{C}_{\mathrm{r}}=45\) \({ }^n \mathrm{C}_{\mathrm{r}+1}=120\) Use of \(\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\) Divide equation (ii) by equation (i) \(\quad \frac{{ }^{\mathrm{n}} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}-1}}=\frac{45}{10} \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{9}{2}\) \(\Rightarrow \mathrm{n}-\mathrm{r}+1=\frac{9}{2} \mathrm{r}\) Divide equation (iii)by equation (ii) \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{120}{45} \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{8}{3}\) \(\Rightarrow \mathrm{n}-\mathrm{r}=\frac{8}{3}(\mathrm{r}+1)\) Therefore, \(\frac{8}{3}(\mathrm{r}+1)+1=\frac{9}{2} \mathrm{r}\) \(\Rightarrow \frac{8 \mathrm{r}+8+3}{3}=\frac{9 \mathrm{r}}{2}\) \(\Rightarrow 16 \mathrm{r}+22=27 \mathrm{r}\) \(\Rightarrow 11 \mathrm{r}=22\) \(\Rightarrow \mathrm{r}=2\)
BCECE-2013
Permutation and Combination
119271
The number of triplets \((x, y, z)\). where \(x, y, z\) are distinct non negative integers satisfying \(\mathbf{x}+\mathbf{y}+\mathrm{z}=\mathbf{1 5}\) is
1 80
2 114
3 92
4 136
Explanation:
B Given that, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=15\) Total no. of solution \(={ }^{15+3-1} \mathrm{C}_{3-1}=136\) \(\operatorname{Le} x=y \neq z\) \(2 x+z=15 \Rightarrow z=15-2 x\) \(r \in\{0,, 2, \ldots 7\}-\{5\}\) \(\therefore 7 \text { solutions }\) \(\therefore \text { there are } 21 \text { solutions in which exactly }\) \(\text { Two of } x, y, z \text { are equal ..(ii) }\) \(\text { There is one solution in which } x=y=z\) \(\text { Required answer }=136-21-1=114\)
119267
\(P(n)=n(n+1)(n+5), n \in N\) is a multiple of
1 9
2 5
3 3
4 7
Explanation:
C : Here, \(p(n)=n(n+1)(n+5)\) \(P(1)=1(2)(6)=12=4 \times 3\) \(P(2)=2(3)(7)=42=14 \times 3\) \(P(3)=3(4)(8)=96=32 \times 3\)3 is common multiple in all \(n \in \mathrm{N}\).
JCECE-2019
Permutation and Combination
119268
The value of \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is
1 \(\frac{2^{50}}{51}\)
2 \(\frac{2^{50}-1}{51}\)
3 \(\frac{2^{50}-1}{50}\)
4 None of these
Explanation:
A Given that, \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots . .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is \(=\left(\frac{50 !}{\frac{50 !}{1}}+\frac{50 !}{\frac{2 ! 48 !}{3}}+\ldots \ldots \frac{50 !}{\frac{50 ! 0 !}{51}}\right)\) \(=\left(1+\frac{50 \times 49}{3 \times 2 !}+\ldots \ldots .+\frac{1}{51}\right)\) Multiply the above expression by \(\frac{51}{51}\) \(=\frac{1}{51}\left(51+\frac{51 \times 50 \times 99}{3 !}+\ldots . .\right)\) \(=\frac{1}{51}\left({ }^{51} \mathrm{C}_1+{ }^{51} \mathrm{C}_3+\ldots .\right)\) As, we know that sum of odd coefficient \(=2^{\mathrm{n}-1}\) \(=\frac{1}{51} 2^{51-1}=\frac{2^{50}}{51}\)
BCECE-2014
Permutation and Combination
119269
If \({ }^n C_{r-1}=10,{ }^n C_r=45\), and \({ }^n C_{r+1}=120\), then \(r\) equals to
1 1
2 2
3 3
4 4
Explanation:
B : We have, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=10\) \({ }^n \mathrm{C}_{\mathrm{r}}=45\) \({ }^n \mathrm{C}_{\mathrm{r}+1}=120\) Use of \(\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\) Divide equation (ii) by equation (i) \(\quad \frac{{ }^{\mathrm{n}} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}-1}}=\frac{45}{10} \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{9}{2}\) \(\Rightarrow \mathrm{n}-\mathrm{r}+1=\frac{9}{2} \mathrm{r}\) Divide equation (iii)by equation (ii) \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{120}{45} \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{8}{3}\) \(\Rightarrow \mathrm{n}-\mathrm{r}=\frac{8}{3}(\mathrm{r}+1)\) Therefore, \(\frac{8}{3}(\mathrm{r}+1)+1=\frac{9}{2} \mathrm{r}\) \(\Rightarrow \frac{8 \mathrm{r}+8+3}{3}=\frac{9 \mathrm{r}}{2}\) \(\Rightarrow 16 \mathrm{r}+22=27 \mathrm{r}\) \(\Rightarrow 11 \mathrm{r}=22\) \(\Rightarrow \mathrm{r}=2\)
BCECE-2013
Permutation and Combination
119271
The number of triplets \((x, y, z)\). where \(x, y, z\) are distinct non negative integers satisfying \(\mathbf{x}+\mathbf{y}+\mathrm{z}=\mathbf{1 5}\) is
1 80
2 114
3 92
4 136
Explanation:
B Given that, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=15\) Total no. of solution \(={ }^{15+3-1} \mathrm{C}_{3-1}=136\) \(\operatorname{Le} x=y \neq z\) \(2 x+z=15 \Rightarrow z=15-2 x\) \(r \in\{0,, 2, \ldots 7\}-\{5\}\) \(\therefore 7 \text { solutions }\) \(\therefore \text { there are } 21 \text { solutions in which exactly }\) \(\text { Two of } x, y, z \text { are equal ..(ii) }\) \(\text { There is one solution in which } x=y=z\) \(\text { Required answer }=136-21-1=114\)
119267
\(P(n)=n(n+1)(n+5), n \in N\) is a multiple of
1 9
2 5
3 3
4 7
Explanation:
C : Here, \(p(n)=n(n+1)(n+5)\) \(P(1)=1(2)(6)=12=4 \times 3\) \(P(2)=2(3)(7)=42=14 \times 3\) \(P(3)=3(4)(8)=96=32 \times 3\)3 is common multiple in all \(n \in \mathrm{N}\).
JCECE-2019
Permutation and Combination
119268
The value of \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is
1 \(\frac{2^{50}}{51}\)
2 \(\frac{2^{50}-1}{51}\)
3 \(\frac{2^{50}-1}{50}\)
4 None of these
Explanation:
A Given that, \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots . .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is \(=\left(\frac{50 !}{\frac{50 !}{1}}+\frac{50 !}{\frac{2 ! 48 !}{3}}+\ldots \ldots \frac{50 !}{\frac{50 ! 0 !}{51}}\right)\) \(=\left(1+\frac{50 \times 49}{3 \times 2 !}+\ldots \ldots .+\frac{1}{51}\right)\) Multiply the above expression by \(\frac{51}{51}\) \(=\frac{1}{51}\left(51+\frac{51 \times 50 \times 99}{3 !}+\ldots . .\right)\) \(=\frac{1}{51}\left({ }^{51} \mathrm{C}_1+{ }^{51} \mathrm{C}_3+\ldots .\right)\) As, we know that sum of odd coefficient \(=2^{\mathrm{n}-1}\) \(=\frac{1}{51} 2^{51-1}=\frac{2^{50}}{51}\)
BCECE-2014
Permutation and Combination
119269
If \({ }^n C_{r-1}=10,{ }^n C_r=45\), and \({ }^n C_{r+1}=120\), then \(r\) equals to
1 1
2 2
3 3
4 4
Explanation:
B : We have, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=10\) \({ }^n \mathrm{C}_{\mathrm{r}}=45\) \({ }^n \mathrm{C}_{\mathrm{r}+1}=120\) Use of \(\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\) Divide equation (ii) by equation (i) \(\quad \frac{{ }^{\mathrm{n}} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}-1}}=\frac{45}{10} \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{9}{2}\) \(\Rightarrow \mathrm{n}-\mathrm{r}+1=\frac{9}{2} \mathrm{r}\) Divide equation (iii)by equation (ii) \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{120}{45} \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{8}{3}\) \(\Rightarrow \mathrm{n}-\mathrm{r}=\frac{8}{3}(\mathrm{r}+1)\) Therefore, \(\frac{8}{3}(\mathrm{r}+1)+1=\frac{9}{2} \mathrm{r}\) \(\Rightarrow \frac{8 \mathrm{r}+8+3}{3}=\frac{9 \mathrm{r}}{2}\) \(\Rightarrow 16 \mathrm{r}+22=27 \mathrm{r}\) \(\Rightarrow 11 \mathrm{r}=22\) \(\Rightarrow \mathrm{r}=2\)
BCECE-2013
Permutation and Combination
119271
The number of triplets \((x, y, z)\). where \(x, y, z\) are distinct non negative integers satisfying \(\mathbf{x}+\mathbf{y}+\mathrm{z}=\mathbf{1 5}\) is
1 80
2 114
3 92
4 136
Explanation:
B Given that, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=15\) Total no. of solution \(={ }^{15+3-1} \mathrm{C}_{3-1}=136\) \(\operatorname{Le} x=y \neq z\) \(2 x+z=15 \Rightarrow z=15-2 x\) \(r \in\{0,, 2, \ldots 7\}-\{5\}\) \(\therefore 7 \text { solutions }\) \(\therefore \text { there are } 21 \text { solutions in which exactly }\) \(\text { Two of } x, y, z \text { are equal ..(ii) }\) \(\text { There is one solution in which } x=y=z\) \(\text { Required answer }=136-21-1=114\)
119267
\(P(n)=n(n+1)(n+5), n \in N\) is a multiple of
1 9
2 5
3 3
4 7
Explanation:
C : Here, \(p(n)=n(n+1)(n+5)\) \(P(1)=1(2)(6)=12=4 \times 3\) \(P(2)=2(3)(7)=42=14 \times 3\) \(P(3)=3(4)(8)=96=32 \times 3\)3 is common multiple in all \(n \in \mathrm{N}\).
JCECE-2019
Permutation and Combination
119268
The value of \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is
1 \(\frac{2^{50}}{51}\)
2 \(\frac{2^{50}-1}{51}\)
3 \(\frac{2^{50}-1}{50}\)
4 None of these
Explanation:
A Given that, \(\left(\frac{{ }^{50} \mathrm{C}_0}{1}+\frac{{ }^{50} \mathrm{C}_2}{3}+\ldots . .+\frac{{ }^{50} \mathrm{C}_{50}}{51}\right)\) is \(=\left(\frac{50 !}{\frac{50 !}{1}}+\frac{50 !}{\frac{2 ! 48 !}{3}}+\ldots \ldots \frac{50 !}{\frac{50 ! 0 !}{51}}\right)\) \(=\left(1+\frac{50 \times 49}{3 \times 2 !}+\ldots \ldots .+\frac{1}{51}\right)\) Multiply the above expression by \(\frac{51}{51}\) \(=\frac{1}{51}\left(51+\frac{51 \times 50 \times 99}{3 !}+\ldots . .\right)\) \(=\frac{1}{51}\left({ }^{51} \mathrm{C}_1+{ }^{51} \mathrm{C}_3+\ldots .\right)\) As, we know that sum of odd coefficient \(=2^{\mathrm{n}-1}\) \(=\frac{1}{51} 2^{51-1}=\frac{2^{50}}{51}\)
BCECE-2014
Permutation and Combination
119269
If \({ }^n C_{r-1}=10,{ }^n C_r=45\), and \({ }^n C_{r+1}=120\), then \(r\) equals to
1 1
2 2
3 3
4 4
Explanation:
B : We have, \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}=10\) \({ }^n \mathrm{C}_{\mathrm{r}}=45\) \({ }^n \mathrm{C}_{\mathrm{r}+1}=120\) Use of \(\frac{{ }^n C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\) Divide equation (ii) by equation (i) \(\quad \frac{{ }^{\mathrm{n}} C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}-1}}=\frac{45}{10} \Rightarrow \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{9}{2}\) \(\Rightarrow \mathrm{n}-\mathrm{r}+1=\frac{9}{2} \mathrm{r}\) Divide equation (iii)by equation (ii) \(\frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}+1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{120}{45} \Rightarrow \frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{8}{3}\) \(\Rightarrow \mathrm{n}-\mathrm{r}=\frac{8}{3}(\mathrm{r}+1)\) Therefore, \(\frac{8}{3}(\mathrm{r}+1)+1=\frac{9}{2} \mathrm{r}\) \(\Rightarrow \frac{8 \mathrm{r}+8+3}{3}=\frac{9 \mathrm{r}}{2}\) \(\Rightarrow 16 \mathrm{r}+22=27 \mathrm{r}\) \(\Rightarrow 11 \mathrm{r}=22\) \(\Rightarrow \mathrm{r}=2\)
BCECE-2013
Permutation and Combination
119271
The number of triplets \((x, y, z)\). where \(x, y, z\) are distinct non negative integers satisfying \(\mathbf{x}+\mathbf{y}+\mathrm{z}=\mathbf{1 5}\) is
1 80
2 114
3 92
4 136
Explanation:
B Given that, \(\mathrm{x}+\mathrm{y}+\mathrm{z}=15\) Total no. of solution \(={ }^{15+3-1} \mathrm{C}_{3-1}=136\) \(\operatorname{Le} x=y \neq z\) \(2 x+z=15 \Rightarrow z=15-2 x\) \(r \in\{0,, 2, \ldots 7\}-\{5\}\) \(\therefore 7 \text { solutions }\) \(\therefore \text { there are } 21 \text { solutions in which exactly }\) \(\text { Two of } x, y, z \text { are equal ..(ii) }\) \(\text { There is one solution in which } x=y=z\) \(\text { Required answer }=136-21-1=114\)
