119202
The number of integers, greater that 7000 that can be formed, using the digits \(3,5,6,7,8\) without repetition, is
1 168
2 120
3 220
4 48
Explanation:
A Given digits are 3, 5, 6, 7, 8 Case I: Four digits numbers starts with 7,8 \(=2 \times 4 \times 3 \times 2=48\) Case-II: Five digit numbers \(=5\) ! \(=5 \times 4 \times 3 \times 2 \times 1\) \(=120\) \(\therefore\) Total number \(=48+120\) \(=168\)
JEE Main-24.01.2023
Permutation and Combination
119204
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
1 400
2 472
3 432
4 507
Explanation:
C Total number of three digit \(=900\) Divisible by \(3=300\) \(\left(\because \frac{900}{3}=300\right)\) No. divisible by \(12=75\) No. divisible by \(4=\frac{900}{4}=225\) Number divisible by either 3 or 4 \(=300+225-75=450\) We have to remove divisible by \(48=18\) Required number of numbers \(=450-18=432\)
JEE Main-29.01.2023
Permutation and Combination
119205
Let the number (22) \({ }^{2022}+\) (2022) \(^{22}\) leave the remainder \(\alpha\) when divided by 3 and \(\beta\) when divided by 7 . Then \(\left(\alpha^2+\beta^2\right)\) is equal to
1 10
2 5
3 20
4 13
Explanation:
B \((22)^{2022}+(2022)^{22}\) For \(\alpha\) Divided by 3 \((21+1)^{2022}+(2022)^{22}=3 \mathrm{k}+1\) \(\alpha=1\) And for \(\beta\) divided by 7 \((21+1)^{2022}+(2023-1)^{22}=7 \mathrm{k}+1+1=7 \mathrm{k}+2\) \(\beta=2\) Hence, \(\alpha=1\) and \(\beta=2\) Therefore, \(\alpha^2+\beta^2=1^2+2^2=1+4=5\)
JEE Main-10.04.2023
Permutation and Combination
119208
Which of the following in an incorrect statement?
1 \(n^3+3 n^2+5 n+3\) is divisible by 3 for all \(n \in\)
2 \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) is divisible by 6 for all \(\mathrm{n} \in\) \(\mathrm{N}\)
3 \(\mathrm{n}^2-\mathrm{n}+41\) is a prime number for all \(\mathrm{n} \in \mathrm{N}\)
4 \(7^{\mathrm{n}}-3^{\mathrm{n}}\) in divisible by 4 for all \(\mathrm{n} \in \mathrm{N}\) Where \(\mathrm{N}\) denotes the set of all natural numbers.
Explanation:
C Let \(\mathrm{p}(\mathrm{n})=\mathrm{n}^2-\mathrm{n}+41\) is a prime number \(\mathrm{p}(1)=1^2-1+41=41\) is a prime number \(\mathrm{p}(2)=2^2-2+41 \Rightarrow 43\) which is prime number \(\mathrm{p}(3)=3^2-3+41 \Rightarrow 47\) which is prime number \(\mathrm{p}(4)=4^2-4+41 \Rightarrow 53\) is a prime \(\Rightarrow\) which is true \(p(41)=41 \times 41-41+41=1681\) which is equal to \((41)^2\), a perfect square, not a prime number. Hence, \(p(n)\) is not true for all \(n \in N\)
AMU-2018
Permutation and Combination
119209
A five digit number divisible by 3 is to be formed using the numbers \(0,1,3,4\) and 5 without repetition. The total number of ways this can be done is :
1 216
2 600
3 240
4 3125
Explanation:
A A number is divisible by 3 if the sum of the digits of the number is divisible by 3 . So, we get two pairs \(0,1,2,4,5\) and \(1,2,3,4,5\) When, sum of 5 digit numbers is divisible by 3 Number of ways \(=4 \times 4 !+5\) ! \(=4 \times 24+120=96+120=216\)
119202
The number of integers, greater that 7000 that can be formed, using the digits \(3,5,6,7,8\) without repetition, is
1 168
2 120
3 220
4 48
Explanation:
A Given digits are 3, 5, 6, 7, 8 Case I: Four digits numbers starts with 7,8 \(=2 \times 4 \times 3 \times 2=48\) Case-II: Five digit numbers \(=5\) ! \(=5 \times 4 \times 3 \times 2 \times 1\) \(=120\) \(\therefore\) Total number \(=48+120\) \(=168\)
JEE Main-24.01.2023
Permutation and Combination
119204
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
1 400
2 472
3 432
4 507
Explanation:
C Total number of three digit \(=900\) Divisible by \(3=300\) \(\left(\because \frac{900}{3}=300\right)\) No. divisible by \(12=75\) No. divisible by \(4=\frac{900}{4}=225\) Number divisible by either 3 or 4 \(=300+225-75=450\) We have to remove divisible by \(48=18\) Required number of numbers \(=450-18=432\)
JEE Main-29.01.2023
Permutation and Combination
119205
Let the number (22) \({ }^{2022}+\) (2022) \(^{22}\) leave the remainder \(\alpha\) when divided by 3 and \(\beta\) when divided by 7 . Then \(\left(\alpha^2+\beta^2\right)\) is equal to
1 10
2 5
3 20
4 13
Explanation:
B \((22)^{2022}+(2022)^{22}\) For \(\alpha\) Divided by 3 \((21+1)^{2022}+(2022)^{22}=3 \mathrm{k}+1\) \(\alpha=1\) And for \(\beta\) divided by 7 \((21+1)^{2022}+(2023-1)^{22}=7 \mathrm{k}+1+1=7 \mathrm{k}+2\) \(\beta=2\) Hence, \(\alpha=1\) and \(\beta=2\) Therefore, \(\alpha^2+\beta^2=1^2+2^2=1+4=5\)
JEE Main-10.04.2023
Permutation and Combination
119208
Which of the following in an incorrect statement?
1 \(n^3+3 n^2+5 n+3\) is divisible by 3 for all \(n \in\)
2 \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) is divisible by 6 for all \(\mathrm{n} \in\) \(\mathrm{N}\)
3 \(\mathrm{n}^2-\mathrm{n}+41\) is a prime number for all \(\mathrm{n} \in \mathrm{N}\)
4 \(7^{\mathrm{n}}-3^{\mathrm{n}}\) in divisible by 4 for all \(\mathrm{n} \in \mathrm{N}\) Where \(\mathrm{N}\) denotes the set of all natural numbers.
Explanation:
C Let \(\mathrm{p}(\mathrm{n})=\mathrm{n}^2-\mathrm{n}+41\) is a prime number \(\mathrm{p}(1)=1^2-1+41=41\) is a prime number \(\mathrm{p}(2)=2^2-2+41 \Rightarrow 43\) which is prime number \(\mathrm{p}(3)=3^2-3+41 \Rightarrow 47\) which is prime number \(\mathrm{p}(4)=4^2-4+41 \Rightarrow 53\) is a prime \(\Rightarrow\) which is true \(p(41)=41 \times 41-41+41=1681\) which is equal to \((41)^2\), a perfect square, not a prime number. Hence, \(p(n)\) is not true for all \(n \in N\)
AMU-2018
Permutation and Combination
119209
A five digit number divisible by 3 is to be formed using the numbers \(0,1,3,4\) and 5 without repetition. The total number of ways this can be done is :
1 216
2 600
3 240
4 3125
Explanation:
A A number is divisible by 3 if the sum of the digits of the number is divisible by 3 . So, we get two pairs \(0,1,2,4,5\) and \(1,2,3,4,5\) When, sum of 5 digit numbers is divisible by 3 Number of ways \(=4 \times 4 !+5\) ! \(=4 \times 24+120=96+120=216\)
119202
The number of integers, greater that 7000 that can be formed, using the digits \(3,5,6,7,8\) without repetition, is
1 168
2 120
3 220
4 48
Explanation:
A Given digits are 3, 5, 6, 7, 8 Case I: Four digits numbers starts with 7,8 \(=2 \times 4 \times 3 \times 2=48\) Case-II: Five digit numbers \(=5\) ! \(=5 \times 4 \times 3 \times 2 \times 1\) \(=120\) \(\therefore\) Total number \(=48+120\) \(=168\)
JEE Main-24.01.2023
Permutation and Combination
119204
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
1 400
2 472
3 432
4 507
Explanation:
C Total number of three digit \(=900\) Divisible by \(3=300\) \(\left(\because \frac{900}{3}=300\right)\) No. divisible by \(12=75\) No. divisible by \(4=\frac{900}{4}=225\) Number divisible by either 3 or 4 \(=300+225-75=450\) We have to remove divisible by \(48=18\) Required number of numbers \(=450-18=432\)
JEE Main-29.01.2023
Permutation and Combination
119205
Let the number (22) \({ }^{2022}+\) (2022) \(^{22}\) leave the remainder \(\alpha\) when divided by 3 and \(\beta\) when divided by 7 . Then \(\left(\alpha^2+\beta^2\right)\) is equal to
1 10
2 5
3 20
4 13
Explanation:
B \((22)^{2022}+(2022)^{22}\) For \(\alpha\) Divided by 3 \((21+1)^{2022}+(2022)^{22}=3 \mathrm{k}+1\) \(\alpha=1\) And for \(\beta\) divided by 7 \((21+1)^{2022}+(2023-1)^{22}=7 \mathrm{k}+1+1=7 \mathrm{k}+2\) \(\beta=2\) Hence, \(\alpha=1\) and \(\beta=2\) Therefore, \(\alpha^2+\beta^2=1^2+2^2=1+4=5\)
JEE Main-10.04.2023
Permutation and Combination
119208
Which of the following in an incorrect statement?
1 \(n^3+3 n^2+5 n+3\) is divisible by 3 for all \(n \in\)
2 \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) is divisible by 6 for all \(\mathrm{n} \in\) \(\mathrm{N}\)
3 \(\mathrm{n}^2-\mathrm{n}+41\) is a prime number for all \(\mathrm{n} \in \mathrm{N}\)
4 \(7^{\mathrm{n}}-3^{\mathrm{n}}\) in divisible by 4 for all \(\mathrm{n} \in \mathrm{N}\) Where \(\mathrm{N}\) denotes the set of all natural numbers.
Explanation:
C Let \(\mathrm{p}(\mathrm{n})=\mathrm{n}^2-\mathrm{n}+41\) is a prime number \(\mathrm{p}(1)=1^2-1+41=41\) is a prime number \(\mathrm{p}(2)=2^2-2+41 \Rightarrow 43\) which is prime number \(\mathrm{p}(3)=3^2-3+41 \Rightarrow 47\) which is prime number \(\mathrm{p}(4)=4^2-4+41 \Rightarrow 53\) is a prime \(\Rightarrow\) which is true \(p(41)=41 \times 41-41+41=1681\) which is equal to \((41)^2\), a perfect square, not a prime number. Hence, \(p(n)\) is not true for all \(n \in N\)
AMU-2018
Permutation and Combination
119209
A five digit number divisible by 3 is to be formed using the numbers \(0,1,3,4\) and 5 without repetition. The total number of ways this can be done is :
1 216
2 600
3 240
4 3125
Explanation:
A A number is divisible by 3 if the sum of the digits of the number is divisible by 3 . So, we get two pairs \(0,1,2,4,5\) and \(1,2,3,4,5\) When, sum of 5 digit numbers is divisible by 3 Number of ways \(=4 \times 4 !+5\) ! \(=4 \times 24+120=96+120=216\)
119202
The number of integers, greater that 7000 that can be formed, using the digits \(3,5,6,7,8\) without repetition, is
1 168
2 120
3 220
4 48
Explanation:
A Given digits are 3, 5, 6, 7, 8 Case I: Four digits numbers starts with 7,8 \(=2 \times 4 \times 3 \times 2=48\) Case-II: Five digit numbers \(=5\) ! \(=5 \times 4 \times 3 \times 2 \times 1\) \(=120\) \(\therefore\) Total number \(=48+120\) \(=168\)
JEE Main-24.01.2023
Permutation and Combination
119204
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
1 400
2 472
3 432
4 507
Explanation:
C Total number of three digit \(=900\) Divisible by \(3=300\) \(\left(\because \frac{900}{3}=300\right)\) No. divisible by \(12=75\) No. divisible by \(4=\frac{900}{4}=225\) Number divisible by either 3 or 4 \(=300+225-75=450\) We have to remove divisible by \(48=18\) Required number of numbers \(=450-18=432\)
JEE Main-29.01.2023
Permutation and Combination
119205
Let the number (22) \({ }^{2022}+\) (2022) \(^{22}\) leave the remainder \(\alpha\) when divided by 3 and \(\beta\) when divided by 7 . Then \(\left(\alpha^2+\beta^2\right)\) is equal to
1 10
2 5
3 20
4 13
Explanation:
B \((22)^{2022}+(2022)^{22}\) For \(\alpha\) Divided by 3 \((21+1)^{2022}+(2022)^{22}=3 \mathrm{k}+1\) \(\alpha=1\) And for \(\beta\) divided by 7 \((21+1)^{2022}+(2023-1)^{22}=7 \mathrm{k}+1+1=7 \mathrm{k}+2\) \(\beta=2\) Hence, \(\alpha=1\) and \(\beta=2\) Therefore, \(\alpha^2+\beta^2=1^2+2^2=1+4=5\)
JEE Main-10.04.2023
Permutation and Combination
119208
Which of the following in an incorrect statement?
1 \(n^3+3 n^2+5 n+3\) is divisible by 3 for all \(n \in\)
2 \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) is divisible by 6 for all \(\mathrm{n} \in\) \(\mathrm{N}\)
3 \(\mathrm{n}^2-\mathrm{n}+41\) is a prime number for all \(\mathrm{n} \in \mathrm{N}\)
4 \(7^{\mathrm{n}}-3^{\mathrm{n}}\) in divisible by 4 for all \(\mathrm{n} \in \mathrm{N}\) Where \(\mathrm{N}\) denotes the set of all natural numbers.
Explanation:
C Let \(\mathrm{p}(\mathrm{n})=\mathrm{n}^2-\mathrm{n}+41\) is a prime number \(\mathrm{p}(1)=1^2-1+41=41\) is a prime number \(\mathrm{p}(2)=2^2-2+41 \Rightarrow 43\) which is prime number \(\mathrm{p}(3)=3^2-3+41 \Rightarrow 47\) which is prime number \(\mathrm{p}(4)=4^2-4+41 \Rightarrow 53\) is a prime \(\Rightarrow\) which is true \(p(41)=41 \times 41-41+41=1681\) which is equal to \((41)^2\), a perfect square, not a prime number. Hence, \(p(n)\) is not true for all \(n \in N\)
AMU-2018
Permutation and Combination
119209
A five digit number divisible by 3 is to be formed using the numbers \(0,1,3,4\) and 5 without repetition. The total number of ways this can be done is :
1 216
2 600
3 240
4 3125
Explanation:
A A number is divisible by 3 if the sum of the digits of the number is divisible by 3 . So, we get two pairs \(0,1,2,4,5\) and \(1,2,3,4,5\) When, sum of 5 digit numbers is divisible by 3 Number of ways \(=4 \times 4 !+5\) ! \(=4 \times 24+120=96+120=216\)
119202
The number of integers, greater that 7000 that can be formed, using the digits \(3,5,6,7,8\) without repetition, is
1 168
2 120
3 220
4 48
Explanation:
A Given digits are 3, 5, 6, 7, 8 Case I: Four digits numbers starts with 7,8 \(=2 \times 4 \times 3 \times 2=48\) Case-II: Five digit numbers \(=5\) ! \(=5 \times 4 \times 3 \times 2 \times 1\) \(=120\) \(\therefore\) Total number \(=48+120\) \(=168\)
JEE Main-24.01.2023
Permutation and Combination
119204
The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48 , is
1 400
2 472
3 432
4 507
Explanation:
C Total number of three digit \(=900\) Divisible by \(3=300\) \(\left(\because \frac{900}{3}=300\right)\) No. divisible by \(12=75\) No. divisible by \(4=\frac{900}{4}=225\) Number divisible by either 3 or 4 \(=300+225-75=450\) We have to remove divisible by \(48=18\) Required number of numbers \(=450-18=432\)
JEE Main-29.01.2023
Permutation and Combination
119205
Let the number (22) \({ }^{2022}+\) (2022) \(^{22}\) leave the remainder \(\alpha\) when divided by 3 and \(\beta\) when divided by 7 . Then \(\left(\alpha^2+\beta^2\right)\) is equal to
1 10
2 5
3 20
4 13
Explanation:
B \((22)^{2022}+(2022)^{22}\) For \(\alpha\) Divided by 3 \((21+1)^{2022}+(2022)^{22}=3 \mathrm{k}+1\) \(\alpha=1\) And for \(\beta\) divided by 7 \((21+1)^{2022}+(2023-1)^{22}=7 \mathrm{k}+1+1=7 \mathrm{k}+2\) \(\beta=2\) Hence, \(\alpha=1\) and \(\beta=2\) Therefore, \(\alpha^2+\beta^2=1^2+2^2=1+4=5\)
JEE Main-10.04.2023
Permutation and Combination
119208
Which of the following in an incorrect statement?
1 \(n^3+3 n^2+5 n+3\) is divisible by 3 for all \(n \in\)
2 \(\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\) is divisible by 6 for all \(\mathrm{n} \in\) \(\mathrm{N}\)
3 \(\mathrm{n}^2-\mathrm{n}+41\) is a prime number for all \(\mathrm{n} \in \mathrm{N}\)
4 \(7^{\mathrm{n}}-3^{\mathrm{n}}\) in divisible by 4 for all \(\mathrm{n} \in \mathrm{N}\) Where \(\mathrm{N}\) denotes the set of all natural numbers.
Explanation:
C Let \(\mathrm{p}(\mathrm{n})=\mathrm{n}^2-\mathrm{n}+41\) is a prime number \(\mathrm{p}(1)=1^2-1+41=41\) is a prime number \(\mathrm{p}(2)=2^2-2+41 \Rightarrow 43\) which is prime number \(\mathrm{p}(3)=3^2-3+41 \Rightarrow 47\) which is prime number \(\mathrm{p}(4)=4^2-4+41 \Rightarrow 53\) is a prime \(\Rightarrow\) which is true \(p(41)=41 \times 41-41+41=1681\) which is equal to \((41)^2\), a perfect square, not a prime number. Hence, \(p(n)\) is not true for all \(n \in N\)
AMU-2018
Permutation and Combination
119209
A five digit number divisible by 3 is to be formed using the numbers \(0,1,3,4\) and 5 without repetition. The total number of ways this can be done is :
1 216
2 600
3 240
4 3125
Explanation:
A A number is divisible by 3 if the sum of the digits of the number is divisible by 3 . So, we get two pairs \(0,1,2,4,5\) and \(1,2,3,4,5\) When, sum of 5 digit numbers is divisible by 3 Number of ways \(=4 \times 4 !+5\) ! \(=4 \times 24+120=96+120=216\)
