119306
Using the letters of the word TRICK a five letter word with distinct letters is formed such that \(C\) is in the middle. In how many ways this is possible?
1 6
2 120
3 24
4 72
Explanation:
C Given the letters of the Word T R I C k Total words = 5 \(=4 !\) \(=4 \times 3 \times 2 \times 1=24\) Total no of ways \(=24\)
TS EAMCET-2017
Permutation and Combination
119308
Among the statements : \(\left(\mathbf{S}_1\right): 2023^{2022}-1999^{2022}\) is divisible by 8 \(\left(\mathbf{S}_{\mathbf{2}}\right): 13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) is divisible by 8 Infinitely many \(\mathbf{n} \in \mathbf{N}\).
1 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are incorrect
2 only \(\left(\mathrm{S}_2\right)\) is correct
3 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are correct
4 only \(\left(\mathrm{S}_1\right)\) is correct
Explanation:
C Statements (1) Given, \(\mathrm{S}_1=2023^{2022}-1999^{2022}\) \(\mathrm{~S}_1=(1999+24)^{2022}-(1999)^{2022}\) \({ }^{2022} \mathrm{C}_1(1999)^{2021}(24)+{ }^{2022} \mathrm{C}_2(1999)^{2020}\) \((24)^2+\ldots \ldots . . \text { Soon }\) \(\mathrm{S}_1\) is divisible by 8 Hence statement \(\mathrm{I}^{\mathrm{st}}\) is correct \(\mathrm{S}_2=13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) \(13^{\mathrm{n}}=(1+12)^{\mathrm{n}}=1+12 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\) \(13\left(13^{\mathrm{n}}\right)-11 \mathrm{n}-13=145 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\)\(\because\) If \((n=144 m, m \in N)\) then If is divisible by 144
JEE Main-06.04.2023
Permutation and Combination
119309
\(\sum_{\mathrm{r}=1}^{20}\left(\mathrm{r}^2+1\right)(\mathrm{r} !)\) is equal to:
NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
119306
Using the letters of the word TRICK a five letter word with distinct letters is formed such that \(C\) is in the middle. In how many ways this is possible?
1 6
2 120
3 24
4 72
Explanation:
C Given the letters of the Word T R I C k Total words = 5 \(=4 !\) \(=4 \times 3 \times 2 \times 1=24\) Total no of ways \(=24\)
TS EAMCET-2017
Permutation and Combination
119308
Among the statements : \(\left(\mathbf{S}_1\right): 2023^{2022}-1999^{2022}\) is divisible by 8 \(\left(\mathbf{S}_{\mathbf{2}}\right): 13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) is divisible by 8 Infinitely many \(\mathbf{n} \in \mathbf{N}\).
1 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are incorrect
2 only \(\left(\mathrm{S}_2\right)\) is correct
3 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are correct
4 only \(\left(\mathrm{S}_1\right)\) is correct
Explanation:
C Statements (1) Given, \(\mathrm{S}_1=2023^{2022}-1999^{2022}\) \(\mathrm{~S}_1=(1999+24)^{2022}-(1999)^{2022}\) \({ }^{2022} \mathrm{C}_1(1999)^{2021}(24)+{ }^{2022} \mathrm{C}_2(1999)^{2020}\) \((24)^2+\ldots \ldots . . \text { Soon }\) \(\mathrm{S}_1\) is divisible by 8 Hence statement \(\mathrm{I}^{\mathrm{st}}\) is correct \(\mathrm{S}_2=13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) \(13^{\mathrm{n}}=(1+12)^{\mathrm{n}}=1+12 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\) \(13\left(13^{\mathrm{n}}\right)-11 \mathrm{n}-13=145 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\)\(\because\) If \((n=144 m, m \in N)\) then If is divisible by 144
JEE Main-06.04.2023
Permutation and Combination
119309
\(\sum_{\mathrm{r}=1}^{20}\left(\mathrm{r}^2+1\right)(\mathrm{r} !)\) is equal to:
119306
Using the letters of the word TRICK a five letter word with distinct letters is formed such that \(C\) is in the middle. In how many ways this is possible?
1 6
2 120
3 24
4 72
Explanation:
C Given the letters of the Word T R I C k Total words = 5 \(=4 !\) \(=4 \times 3 \times 2 \times 1=24\) Total no of ways \(=24\)
TS EAMCET-2017
Permutation and Combination
119308
Among the statements : \(\left(\mathbf{S}_1\right): 2023^{2022}-1999^{2022}\) is divisible by 8 \(\left(\mathbf{S}_{\mathbf{2}}\right): 13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) is divisible by 8 Infinitely many \(\mathbf{n} \in \mathbf{N}\).
1 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are incorrect
2 only \(\left(\mathrm{S}_2\right)\) is correct
3 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are correct
4 only \(\left(\mathrm{S}_1\right)\) is correct
Explanation:
C Statements (1) Given, \(\mathrm{S}_1=2023^{2022}-1999^{2022}\) \(\mathrm{~S}_1=(1999+24)^{2022}-(1999)^{2022}\) \({ }^{2022} \mathrm{C}_1(1999)^{2021}(24)+{ }^{2022} \mathrm{C}_2(1999)^{2020}\) \((24)^2+\ldots \ldots . . \text { Soon }\) \(\mathrm{S}_1\) is divisible by 8 Hence statement \(\mathrm{I}^{\mathrm{st}}\) is correct \(\mathrm{S}_2=13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) \(13^{\mathrm{n}}=(1+12)^{\mathrm{n}}=1+12 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\) \(13\left(13^{\mathrm{n}}\right)-11 \mathrm{n}-13=145 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\)\(\because\) If \((n=144 m, m \in N)\) then If is divisible by 144
JEE Main-06.04.2023
Permutation and Combination
119309
\(\sum_{\mathrm{r}=1}^{20}\left(\mathrm{r}^2+1\right)(\mathrm{r} !)\) is equal to:
119306
Using the letters of the word TRICK a five letter word with distinct letters is formed such that \(C\) is in the middle. In how many ways this is possible?
1 6
2 120
3 24
4 72
Explanation:
C Given the letters of the Word T R I C k Total words = 5 \(=4 !\) \(=4 \times 3 \times 2 \times 1=24\) Total no of ways \(=24\)
TS EAMCET-2017
Permutation and Combination
119308
Among the statements : \(\left(\mathbf{S}_1\right): 2023^{2022}-1999^{2022}\) is divisible by 8 \(\left(\mathbf{S}_{\mathbf{2}}\right): 13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) is divisible by 8 Infinitely many \(\mathbf{n} \in \mathbf{N}\).
1 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are incorrect
2 only \(\left(\mathrm{S}_2\right)\) is correct
3 both \(\left(\mathrm{S}_1\right)\) and \(\left(\mathrm{S}_2\right)\) are correct
4 only \(\left(\mathrm{S}_1\right)\) is correct
Explanation:
C Statements (1) Given, \(\mathrm{S}_1=2023^{2022}-1999^{2022}\) \(\mathrm{~S}_1=(1999+24)^{2022}-(1999)^{2022}\) \({ }^{2022} \mathrm{C}_1(1999)^{2021}(24)+{ }^{2022} \mathrm{C}_2(1999)^{2020}\) \((24)^2+\ldots \ldots . . \text { Soon }\) \(\mathrm{S}_1\) is divisible by 8 Hence statement \(\mathrm{I}^{\mathrm{st}}\) is correct \(\mathrm{S}_2=13(13)^{\mathrm{n}}-11 \mathrm{n}-13\) \(13^{\mathrm{n}}=(1+12)^{\mathrm{n}}=1+12 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\) \(13\left(13^{\mathrm{n}}\right)-11 \mathrm{n}-13=145 \mathrm{n}+{ }^{\mathrm{n}} \mathrm{C}_2 \cdot 12^2+{ }^{\mathrm{n}} \mathrm{C}_3 \cdot 12^3\)\(\because\) If \((n=144 m, m \in N)\) then If is divisible by 144
JEE Main-06.04.2023
Permutation and Combination
119309
\(\sum_{\mathrm{r}=1}^{20}\left(\mathrm{r}^2+1\right)(\mathrm{r} !)\) is equal to:
