119206
The product \(r\) consecutive integers is divisible by
1 \(r\) !
2 \((\mathrm{r}-1)\) !
3 \((\mathrm{r}+1)\) !
4 none of these
Explanation:
A Let the consecutive integers be \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}-1)\) Product \(=n(n+1)(n+2) \ldots(n+r-1)\) \(=\frac{(\mathrm{n}-1) ! \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots \ldots(\mathrm{n}+\mathrm{r}-1) \mathrm{r} !}{(\mathrm{n}-1) ! \mathrm{r} !}\) \(={ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}} \mathrm{r}\) ! \(\therefore\) Product is divisible by \(\mathrm{r}\) !
AMU-2010
Permutation and Combination
119187
How many integers between 200 and 700 consist of three distinct digits?
1 350
2 360
3 365
4 370
Explanation:
B Lowest number with 3 distinct digits is 201 Highest number \(=698\) Now the digit we have \(0,1,2,3,4,5,6,7,8,9\), First place can be filled in 5 ways \((2,3,4,5,6)\) Second place in 9 ways and third place in 8 ways. \(\therefore\) Total number of ways \(=5 \cdot 8 \cdot 9=360\)
MHT CET-2021
Permutation and Combination
119189
The number of positive integers formed with almost 10 digits using the digits 0,12 is
1 59048
2 57512
3 56011
4 7431
Explanation:
A Here, each of the 10 digits can assume any of the three values. The numbers so formed include all zeros Since 0 is not positive integer. So, 0 has to be neglected Therefore, the desired number is \(3^{10}-1\) \(=59049-1=59048\)
COMEDK-2016
Permutation and Combination
119190
The number of five digit numbers which are divisible by 4 that can be formed from the digits \(0,1,2,3,4\), is
1 125
2 30
3 40
4 25
Explanation:
B For the number to be divisible by 4 , last two digit of that number is divisible by 4 . So, the last two digit can be \(04,12,20,24,32,40\) \(\therefore\) The following cases arises; Case (i) Last two digit 04 So, \(3 \times 2 \times 1=6\) ways Case (ii) Last two digit 12 So, \(2 \times 2 \times 1=4\) ways Case (iii) Last two digit 20 So, \(3 \times 2 \times 1=6\) ways Case (iv) Last two digit 24 So, \(2 \times 2 \times 1=4\) ways Case (v) Last two digit 32 So, \(2 \times 2 \times 1=4\) ways Case (vi) Last two digit 40 So, \(3 \times 2 \times 1=6\) ways \(\therefore\) Total numbers formed \(=3(6+4)=30\)
SRM JEEE-2009
Permutation and Combination
119191
5 boys of class VI, 6 boys of class VII and 7 boys of class VIII sit in a row. The number of ways they can sit so that boys of the same class sit together is
1 \((5 !)(6 !)(7 !)\)
2 \((3 !)(5 !)(6 !)(7 !)\)
3 \(18 !(5 ! 6 ! 7 !)\)
4 \((4 !)(5 !)(6 !)\)
Explanation:
B There are three group which can be organized in 3! ways. No. of ways that boys of class VI sit together \(=5\) ! No. of ways that boys of class VII sit together \(=6\) ! No. of ways that boys of class VIII sit together \(=7\) ! \(\therefore\) Total no. of ways \(=3 ! \times 5 ! \times 6 ! \times 7\) !
119206
The product \(r\) consecutive integers is divisible by
1 \(r\) !
2 \((\mathrm{r}-1)\) !
3 \((\mathrm{r}+1)\) !
4 none of these
Explanation:
A Let the consecutive integers be \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}-1)\) Product \(=n(n+1)(n+2) \ldots(n+r-1)\) \(=\frac{(\mathrm{n}-1) ! \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots \ldots(\mathrm{n}+\mathrm{r}-1) \mathrm{r} !}{(\mathrm{n}-1) ! \mathrm{r} !}\) \(={ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}} \mathrm{r}\) ! \(\therefore\) Product is divisible by \(\mathrm{r}\) !
AMU-2010
Permutation and Combination
119187
How many integers between 200 and 700 consist of three distinct digits?
1 350
2 360
3 365
4 370
Explanation:
B Lowest number with 3 distinct digits is 201 Highest number \(=698\) Now the digit we have \(0,1,2,3,4,5,6,7,8,9\), First place can be filled in 5 ways \((2,3,4,5,6)\) Second place in 9 ways and third place in 8 ways. \(\therefore\) Total number of ways \(=5 \cdot 8 \cdot 9=360\)
MHT CET-2021
Permutation and Combination
119189
The number of positive integers formed with almost 10 digits using the digits 0,12 is
1 59048
2 57512
3 56011
4 7431
Explanation:
A Here, each of the 10 digits can assume any of the three values. The numbers so formed include all zeros Since 0 is not positive integer. So, 0 has to be neglected Therefore, the desired number is \(3^{10}-1\) \(=59049-1=59048\)
COMEDK-2016
Permutation and Combination
119190
The number of five digit numbers which are divisible by 4 that can be formed from the digits \(0,1,2,3,4\), is
1 125
2 30
3 40
4 25
Explanation:
B For the number to be divisible by 4 , last two digit of that number is divisible by 4 . So, the last two digit can be \(04,12,20,24,32,40\) \(\therefore\) The following cases arises; Case (i) Last two digit 04 So, \(3 \times 2 \times 1=6\) ways Case (ii) Last two digit 12 So, \(2 \times 2 \times 1=4\) ways Case (iii) Last two digit 20 So, \(3 \times 2 \times 1=6\) ways Case (iv) Last two digit 24 So, \(2 \times 2 \times 1=4\) ways Case (v) Last two digit 32 So, \(2 \times 2 \times 1=4\) ways Case (vi) Last two digit 40 So, \(3 \times 2 \times 1=6\) ways \(\therefore\) Total numbers formed \(=3(6+4)=30\)
SRM JEEE-2009
Permutation and Combination
119191
5 boys of class VI, 6 boys of class VII and 7 boys of class VIII sit in a row. The number of ways they can sit so that boys of the same class sit together is
1 \((5 !)(6 !)(7 !)\)
2 \((3 !)(5 !)(6 !)(7 !)\)
3 \(18 !(5 ! 6 ! 7 !)\)
4 \((4 !)(5 !)(6 !)\)
Explanation:
B There are three group which can be organized in 3! ways. No. of ways that boys of class VI sit together \(=5\) ! No. of ways that boys of class VII sit together \(=6\) ! No. of ways that boys of class VIII sit together \(=7\) ! \(\therefore\) Total no. of ways \(=3 ! \times 5 ! \times 6 ! \times 7\) !
119206
The product \(r\) consecutive integers is divisible by
1 \(r\) !
2 \((\mathrm{r}-1)\) !
3 \((\mathrm{r}+1)\) !
4 none of these
Explanation:
A Let the consecutive integers be \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}-1)\) Product \(=n(n+1)(n+2) \ldots(n+r-1)\) \(=\frac{(\mathrm{n}-1) ! \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots \ldots(\mathrm{n}+\mathrm{r}-1) \mathrm{r} !}{(\mathrm{n}-1) ! \mathrm{r} !}\) \(={ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}} \mathrm{r}\) ! \(\therefore\) Product is divisible by \(\mathrm{r}\) !
AMU-2010
Permutation and Combination
119187
How many integers between 200 and 700 consist of three distinct digits?
1 350
2 360
3 365
4 370
Explanation:
B Lowest number with 3 distinct digits is 201 Highest number \(=698\) Now the digit we have \(0,1,2,3,4,5,6,7,8,9\), First place can be filled in 5 ways \((2,3,4,5,6)\) Second place in 9 ways and third place in 8 ways. \(\therefore\) Total number of ways \(=5 \cdot 8 \cdot 9=360\)
MHT CET-2021
Permutation and Combination
119189
The number of positive integers formed with almost 10 digits using the digits 0,12 is
1 59048
2 57512
3 56011
4 7431
Explanation:
A Here, each of the 10 digits can assume any of the three values. The numbers so formed include all zeros Since 0 is not positive integer. So, 0 has to be neglected Therefore, the desired number is \(3^{10}-1\) \(=59049-1=59048\)
COMEDK-2016
Permutation and Combination
119190
The number of five digit numbers which are divisible by 4 that can be formed from the digits \(0,1,2,3,4\), is
1 125
2 30
3 40
4 25
Explanation:
B For the number to be divisible by 4 , last two digit of that number is divisible by 4 . So, the last two digit can be \(04,12,20,24,32,40\) \(\therefore\) The following cases arises; Case (i) Last two digit 04 So, \(3 \times 2 \times 1=6\) ways Case (ii) Last two digit 12 So, \(2 \times 2 \times 1=4\) ways Case (iii) Last two digit 20 So, \(3 \times 2 \times 1=6\) ways Case (iv) Last two digit 24 So, \(2 \times 2 \times 1=4\) ways Case (v) Last two digit 32 So, \(2 \times 2 \times 1=4\) ways Case (vi) Last two digit 40 So, \(3 \times 2 \times 1=6\) ways \(\therefore\) Total numbers formed \(=3(6+4)=30\)
SRM JEEE-2009
Permutation and Combination
119191
5 boys of class VI, 6 boys of class VII and 7 boys of class VIII sit in a row. The number of ways they can sit so that boys of the same class sit together is
1 \((5 !)(6 !)(7 !)\)
2 \((3 !)(5 !)(6 !)(7 !)\)
3 \(18 !(5 ! 6 ! 7 !)\)
4 \((4 !)(5 !)(6 !)\)
Explanation:
B There are three group which can be organized in 3! ways. No. of ways that boys of class VI sit together \(=5\) ! No. of ways that boys of class VII sit together \(=6\) ! No. of ways that boys of class VIII sit together \(=7\) ! \(\therefore\) Total no. of ways \(=3 ! \times 5 ! \times 6 ! \times 7\) !
119206
The product \(r\) consecutive integers is divisible by
1 \(r\) !
2 \((\mathrm{r}-1)\) !
3 \((\mathrm{r}+1)\) !
4 none of these
Explanation:
A Let the consecutive integers be \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}-1)\) Product \(=n(n+1)(n+2) \ldots(n+r-1)\) \(=\frac{(\mathrm{n}-1) ! \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots \ldots(\mathrm{n}+\mathrm{r}-1) \mathrm{r} !}{(\mathrm{n}-1) ! \mathrm{r} !}\) \(={ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}} \mathrm{r}\) ! \(\therefore\) Product is divisible by \(\mathrm{r}\) !
AMU-2010
Permutation and Combination
119187
How many integers between 200 and 700 consist of three distinct digits?
1 350
2 360
3 365
4 370
Explanation:
B Lowest number with 3 distinct digits is 201 Highest number \(=698\) Now the digit we have \(0,1,2,3,4,5,6,7,8,9\), First place can be filled in 5 ways \((2,3,4,5,6)\) Second place in 9 ways and third place in 8 ways. \(\therefore\) Total number of ways \(=5 \cdot 8 \cdot 9=360\)
MHT CET-2021
Permutation and Combination
119189
The number of positive integers formed with almost 10 digits using the digits 0,12 is
1 59048
2 57512
3 56011
4 7431
Explanation:
A Here, each of the 10 digits can assume any of the three values. The numbers so formed include all zeros Since 0 is not positive integer. So, 0 has to be neglected Therefore, the desired number is \(3^{10}-1\) \(=59049-1=59048\)
COMEDK-2016
Permutation and Combination
119190
The number of five digit numbers which are divisible by 4 that can be formed from the digits \(0,1,2,3,4\), is
1 125
2 30
3 40
4 25
Explanation:
B For the number to be divisible by 4 , last two digit of that number is divisible by 4 . So, the last two digit can be \(04,12,20,24,32,40\) \(\therefore\) The following cases arises; Case (i) Last two digit 04 So, \(3 \times 2 \times 1=6\) ways Case (ii) Last two digit 12 So, \(2 \times 2 \times 1=4\) ways Case (iii) Last two digit 20 So, \(3 \times 2 \times 1=6\) ways Case (iv) Last two digit 24 So, \(2 \times 2 \times 1=4\) ways Case (v) Last two digit 32 So, \(2 \times 2 \times 1=4\) ways Case (vi) Last two digit 40 So, \(3 \times 2 \times 1=6\) ways \(\therefore\) Total numbers formed \(=3(6+4)=30\)
SRM JEEE-2009
Permutation and Combination
119191
5 boys of class VI, 6 boys of class VII and 7 boys of class VIII sit in a row. The number of ways they can sit so that boys of the same class sit together is
1 \((5 !)(6 !)(7 !)\)
2 \((3 !)(5 !)(6 !)(7 !)\)
3 \(18 !(5 ! 6 ! 7 !)\)
4 \((4 !)(5 !)(6 !)\)
Explanation:
B There are three group which can be organized in 3! ways. No. of ways that boys of class VI sit together \(=5\) ! No. of ways that boys of class VII sit together \(=6\) ! No. of ways that boys of class VIII sit together \(=7\) ! \(\therefore\) Total no. of ways \(=3 ! \times 5 ! \times 6 ! \times 7\) !
119206
The product \(r\) consecutive integers is divisible by
1 \(r\) !
2 \((\mathrm{r}-1)\) !
3 \((\mathrm{r}+1)\) !
4 none of these
Explanation:
A Let the consecutive integers be \(\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots(\mathrm{n}+\mathrm{r}-1)\) Product \(=n(n+1)(n+2) \ldots(n+r-1)\) \(=\frac{(\mathrm{n}-1) ! \mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2) \ldots \ldots(\mathrm{n}+\mathrm{r}-1) \mathrm{r} !}{(\mathrm{n}-1) ! \mathrm{r} !}\) \(={ }^{(\mathrm{n}+\mathrm{r}-1)} \mathrm{C}_{\mathrm{r}} \mathrm{r}\) ! \(\therefore\) Product is divisible by \(\mathrm{r}\) !
AMU-2010
Permutation and Combination
119187
How many integers between 200 and 700 consist of three distinct digits?
1 350
2 360
3 365
4 370
Explanation:
B Lowest number with 3 distinct digits is 201 Highest number \(=698\) Now the digit we have \(0,1,2,3,4,5,6,7,8,9\), First place can be filled in 5 ways \((2,3,4,5,6)\) Second place in 9 ways and third place in 8 ways. \(\therefore\) Total number of ways \(=5 \cdot 8 \cdot 9=360\)
MHT CET-2021
Permutation and Combination
119189
The number of positive integers formed with almost 10 digits using the digits 0,12 is
1 59048
2 57512
3 56011
4 7431
Explanation:
A Here, each of the 10 digits can assume any of the three values. The numbers so formed include all zeros Since 0 is not positive integer. So, 0 has to be neglected Therefore, the desired number is \(3^{10}-1\) \(=59049-1=59048\)
COMEDK-2016
Permutation and Combination
119190
The number of five digit numbers which are divisible by 4 that can be formed from the digits \(0,1,2,3,4\), is
1 125
2 30
3 40
4 25
Explanation:
B For the number to be divisible by 4 , last two digit of that number is divisible by 4 . So, the last two digit can be \(04,12,20,24,32,40\) \(\therefore\) The following cases arises; Case (i) Last two digit 04 So, \(3 \times 2 \times 1=6\) ways Case (ii) Last two digit 12 So, \(2 \times 2 \times 1=4\) ways Case (iii) Last two digit 20 So, \(3 \times 2 \times 1=6\) ways Case (iv) Last two digit 24 So, \(2 \times 2 \times 1=4\) ways Case (v) Last two digit 32 So, \(2 \times 2 \times 1=4\) ways Case (vi) Last two digit 40 So, \(3 \times 2 \times 1=6\) ways \(\therefore\) Total numbers formed \(=3(6+4)=30\)
SRM JEEE-2009
Permutation and Combination
119191
5 boys of class VI, 6 boys of class VII and 7 boys of class VIII sit in a row. The number of ways they can sit so that boys of the same class sit together is
1 \((5 !)(6 !)(7 !)\)
2 \((3 !)(5 !)(6 !)(7 !)\)
3 \(18 !(5 ! 6 ! 7 !)\)
4 \((4 !)(5 !)(6 !)\)
Explanation:
B There are three group which can be organized in 3! ways. No. of ways that boys of class VI sit together \(=5\) ! No. of ways that boys of class VII sit together \(=6\) ! No. of ways that boys of class VIII sit together \(=7\) ! \(\therefore\) Total no. of ways \(=3 ! \times 5 ! \times 6 ! \times 7\) !
