118906
The sum of all the 4 digit numbers that can be formed by using the digits \(2,4,6,8\) (repetition of digits is not allowed) is
1 133320
2 123330
3 113230
4 323430
Explanation:
A We have to find the sum of all 4 digit number that can be formed by using the digits \(2,4,6,8\) So, total 4 - digit numbers that can be formed using 2,4 , 6 , and 8 without repetition \(=4 \times 3 \times 2 \times 1=24\) Here, let the sum of unit digits of all 24 number be A. Let the sum of tens digits of all 24 number be \(B\). Let the sum of hundreds digits of all 24 number be \(\mathrm{C}\). Let the sum of thousands digits of all 24 numbers be \(D\). So, \(\left(\mathrm{D} \times 10^3\right)+\left(\mathrm{C} \times 10^2\right)+\left(\mathrm{B} \times 10^1\right)+\left(\mathrm{A} \times 10^0\right)\) We know, each number will come at each place 6 times. \(=\left\{6(2+4+6+8) \times 10^3\right\}+\left\{6(2+4+6+8) \times 10^2\right\}\) \(+\left\{6(2+4+6+8) \times 10^1\right\}+\left\{6(2+4+6+8) \times 10^0\right\}\) \(= 120 \times 10^3+120 \times 10^2+120 \times 10^1+120 \times 10^0\) \(= 12000 0+12000 +1200+120=133320\)
COMEDK-2017
Permutation and Combination
118908
If \({ }^n C_{r-1}=36,{ }^n C_r=84\) and \({ }^n C_{r+1}=126\), then
1 \(\mathrm{r}=3, \mathrm{n}=9\)
2 \(\mathrm{n}=8, \mathrm{r}=4\)
3 \(\mathrm{n}=7, \mathrm{r}=5\)
4 \(\mathrm{r}=6, \mathrm{n}=16\)
Explanation:
A Given that, \({ }^n C_{r-1}=36\) \({ }^n C_r=84\) \({ }^n C_{r+1}=126\) Divide equation (ii) by equation (i), we get \(\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{84}{36}\) \(\Rightarrow \quad \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{7}{3}\) \(\Rightarrow \quad 3 \mathrm{n}-3 \mathrm{r}+3=7 \mathrm{r}\) \(\Rightarrow \quad 3 \mathrm{n}-10 \mathrm{r}+3=0\) Divide equation (iii) by equation (ii), we get :- \(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\) \(\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}\) \(\Rightarrow \quad 2 n-2 r=3 r+3\) \(\Rightarrow \quad 2 n-5 r-3=0\) Solving equation (iv) and (v), we get \(\mathrm{n}=9, \mathrm{r}=3\)
SRM JEEE-2009
Permutation and Combination
118909
If \({ }^n P_r=720{ }^n C_r\), then the value of \(r\) is
1 6
2 5
3 4
4 7
Explanation:
A Given that, \({ }^{{ }^n P_{\mathrm{r}}}=720{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) \(\Rightarrow \quad \frac{n !}{(n-r) !}=720 \times \frac{n !}{(n-r) ! r !}\) \(\mathrm{r} !=720\) \(\Rightarrow \quad \mathrm{r} !=6\) ! \(\Rightarrow \quad \mathrm{r}=6\)
SRM JEEE-2011
Permutation and Combination
118910
The number of positive integral solution of abc \(=30\) is
1 30
2 27
3 8
4 None of these
Explanation:
B Given the positive integral solution, \(\mathrm{abc}=30\) \(\therefore \quad 30=2 \times 3 \times 5\) \(=5 \times 6 \times 1\) \(=15 \times 2 \times 1\) \(=30 \times 1 \times 1\) Here the possibility of 30 is three. It can be as a, b, c and the possibility of 1 can be both of the digits at time. So, this is written as- \(\frac{3 !}{2 !}\) So, number of solution is- \(=3 !+3 !+3 !+3 !+\frac{3 !}{2 !}\) \(=(4 \times 3 !)+\frac{3 !}{2 !}\) \(=(4 \times 6)+3\) \(=24+3\) \(=27\)
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Permutation and Combination
118906
The sum of all the 4 digit numbers that can be formed by using the digits \(2,4,6,8\) (repetition of digits is not allowed) is
1 133320
2 123330
3 113230
4 323430
Explanation:
A We have to find the sum of all 4 digit number that can be formed by using the digits \(2,4,6,8\) So, total 4 - digit numbers that can be formed using 2,4 , 6 , and 8 without repetition \(=4 \times 3 \times 2 \times 1=24\) Here, let the sum of unit digits of all 24 number be A. Let the sum of tens digits of all 24 number be \(B\). Let the sum of hundreds digits of all 24 number be \(\mathrm{C}\). Let the sum of thousands digits of all 24 numbers be \(D\). So, \(\left(\mathrm{D} \times 10^3\right)+\left(\mathrm{C} \times 10^2\right)+\left(\mathrm{B} \times 10^1\right)+\left(\mathrm{A} \times 10^0\right)\) We know, each number will come at each place 6 times. \(=\left\{6(2+4+6+8) \times 10^3\right\}+\left\{6(2+4+6+8) \times 10^2\right\}\) \(+\left\{6(2+4+6+8) \times 10^1\right\}+\left\{6(2+4+6+8) \times 10^0\right\}\) \(= 120 \times 10^3+120 \times 10^2+120 \times 10^1+120 \times 10^0\) \(= 12000 0+12000 +1200+120=133320\)
COMEDK-2017
Permutation and Combination
118908
If \({ }^n C_{r-1}=36,{ }^n C_r=84\) and \({ }^n C_{r+1}=126\), then
1 \(\mathrm{r}=3, \mathrm{n}=9\)
2 \(\mathrm{n}=8, \mathrm{r}=4\)
3 \(\mathrm{n}=7, \mathrm{r}=5\)
4 \(\mathrm{r}=6, \mathrm{n}=16\)
Explanation:
A Given that, \({ }^n C_{r-1}=36\) \({ }^n C_r=84\) \({ }^n C_{r+1}=126\) Divide equation (ii) by equation (i), we get \(\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{84}{36}\) \(\Rightarrow \quad \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{7}{3}\) \(\Rightarrow \quad 3 \mathrm{n}-3 \mathrm{r}+3=7 \mathrm{r}\) \(\Rightarrow \quad 3 \mathrm{n}-10 \mathrm{r}+3=0\) Divide equation (iii) by equation (ii), we get :- \(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\) \(\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}\) \(\Rightarrow \quad 2 n-2 r=3 r+3\) \(\Rightarrow \quad 2 n-5 r-3=0\) Solving equation (iv) and (v), we get \(\mathrm{n}=9, \mathrm{r}=3\)
SRM JEEE-2009
Permutation and Combination
118909
If \({ }^n P_r=720{ }^n C_r\), then the value of \(r\) is
1 6
2 5
3 4
4 7
Explanation:
A Given that, \({ }^{{ }^n P_{\mathrm{r}}}=720{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) \(\Rightarrow \quad \frac{n !}{(n-r) !}=720 \times \frac{n !}{(n-r) ! r !}\) \(\mathrm{r} !=720\) \(\Rightarrow \quad \mathrm{r} !=6\) ! \(\Rightarrow \quad \mathrm{r}=6\)
SRM JEEE-2011
Permutation and Combination
118910
The number of positive integral solution of abc \(=30\) is
1 30
2 27
3 8
4 None of these
Explanation:
B Given the positive integral solution, \(\mathrm{abc}=30\) \(\therefore \quad 30=2 \times 3 \times 5\) \(=5 \times 6 \times 1\) \(=15 \times 2 \times 1\) \(=30 \times 1 \times 1\) Here the possibility of 30 is three. It can be as a, b, c and the possibility of 1 can be both of the digits at time. So, this is written as- \(\frac{3 !}{2 !}\) So, number of solution is- \(=3 !+3 !+3 !+3 !+\frac{3 !}{2 !}\) \(=(4 \times 3 !)+\frac{3 !}{2 !}\) \(=(4 \times 6)+3\) \(=24+3\) \(=27\)
118906
The sum of all the 4 digit numbers that can be formed by using the digits \(2,4,6,8\) (repetition of digits is not allowed) is
1 133320
2 123330
3 113230
4 323430
Explanation:
A We have to find the sum of all 4 digit number that can be formed by using the digits \(2,4,6,8\) So, total 4 - digit numbers that can be formed using 2,4 , 6 , and 8 without repetition \(=4 \times 3 \times 2 \times 1=24\) Here, let the sum of unit digits of all 24 number be A. Let the sum of tens digits of all 24 number be \(B\). Let the sum of hundreds digits of all 24 number be \(\mathrm{C}\). Let the sum of thousands digits of all 24 numbers be \(D\). So, \(\left(\mathrm{D} \times 10^3\right)+\left(\mathrm{C} \times 10^2\right)+\left(\mathrm{B} \times 10^1\right)+\left(\mathrm{A} \times 10^0\right)\) We know, each number will come at each place 6 times. \(=\left\{6(2+4+6+8) \times 10^3\right\}+\left\{6(2+4+6+8) \times 10^2\right\}\) \(+\left\{6(2+4+6+8) \times 10^1\right\}+\left\{6(2+4+6+8) \times 10^0\right\}\) \(= 120 \times 10^3+120 \times 10^2+120 \times 10^1+120 \times 10^0\) \(= 12000 0+12000 +1200+120=133320\)
COMEDK-2017
Permutation and Combination
118908
If \({ }^n C_{r-1}=36,{ }^n C_r=84\) and \({ }^n C_{r+1}=126\), then
1 \(\mathrm{r}=3, \mathrm{n}=9\)
2 \(\mathrm{n}=8, \mathrm{r}=4\)
3 \(\mathrm{n}=7, \mathrm{r}=5\)
4 \(\mathrm{r}=6, \mathrm{n}=16\)
Explanation:
A Given that, \({ }^n C_{r-1}=36\) \({ }^n C_r=84\) \({ }^n C_{r+1}=126\) Divide equation (ii) by equation (i), we get \(\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{84}{36}\) \(\Rightarrow \quad \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{7}{3}\) \(\Rightarrow \quad 3 \mathrm{n}-3 \mathrm{r}+3=7 \mathrm{r}\) \(\Rightarrow \quad 3 \mathrm{n}-10 \mathrm{r}+3=0\) Divide equation (iii) by equation (ii), we get :- \(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\) \(\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}\) \(\Rightarrow \quad 2 n-2 r=3 r+3\) \(\Rightarrow \quad 2 n-5 r-3=0\) Solving equation (iv) and (v), we get \(\mathrm{n}=9, \mathrm{r}=3\)
SRM JEEE-2009
Permutation and Combination
118909
If \({ }^n P_r=720{ }^n C_r\), then the value of \(r\) is
1 6
2 5
3 4
4 7
Explanation:
A Given that, \({ }^{{ }^n P_{\mathrm{r}}}=720{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) \(\Rightarrow \quad \frac{n !}{(n-r) !}=720 \times \frac{n !}{(n-r) ! r !}\) \(\mathrm{r} !=720\) \(\Rightarrow \quad \mathrm{r} !=6\) ! \(\Rightarrow \quad \mathrm{r}=6\)
SRM JEEE-2011
Permutation and Combination
118910
The number of positive integral solution of abc \(=30\) is
1 30
2 27
3 8
4 None of these
Explanation:
B Given the positive integral solution, \(\mathrm{abc}=30\) \(\therefore \quad 30=2 \times 3 \times 5\) \(=5 \times 6 \times 1\) \(=15 \times 2 \times 1\) \(=30 \times 1 \times 1\) Here the possibility of 30 is three. It can be as a, b, c and the possibility of 1 can be both of the digits at time. So, this is written as- \(\frac{3 !}{2 !}\) So, number of solution is- \(=3 !+3 !+3 !+3 !+\frac{3 !}{2 !}\) \(=(4 \times 3 !)+\frac{3 !}{2 !}\) \(=(4 \times 6)+3\) \(=24+3\) \(=27\)
118906
The sum of all the 4 digit numbers that can be formed by using the digits \(2,4,6,8\) (repetition of digits is not allowed) is
1 133320
2 123330
3 113230
4 323430
Explanation:
A We have to find the sum of all 4 digit number that can be formed by using the digits \(2,4,6,8\) So, total 4 - digit numbers that can be formed using 2,4 , 6 , and 8 without repetition \(=4 \times 3 \times 2 \times 1=24\) Here, let the sum of unit digits of all 24 number be A. Let the sum of tens digits of all 24 number be \(B\). Let the sum of hundreds digits of all 24 number be \(\mathrm{C}\). Let the sum of thousands digits of all 24 numbers be \(D\). So, \(\left(\mathrm{D} \times 10^3\right)+\left(\mathrm{C} \times 10^2\right)+\left(\mathrm{B} \times 10^1\right)+\left(\mathrm{A} \times 10^0\right)\) We know, each number will come at each place 6 times. \(=\left\{6(2+4+6+8) \times 10^3\right\}+\left\{6(2+4+6+8) \times 10^2\right\}\) \(+\left\{6(2+4+6+8) \times 10^1\right\}+\left\{6(2+4+6+8) \times 10^0\right\}\) \(= 120 \times 10^3+120 \times 10^2+120 \times 10^1+120 \times 10^0\) \(= 12000 0+12000 +1200+120=133320\)
COMEDK-2017
Permutation and Combination
118908
If \({ }^n C_{r-1}=36,{ }^n C_r=84\) and \({ }^n C_{r+1}=126\), then
1 \(\mathrm{r}=3, \mathrm{n}=9\)
2 \(\mathrm{n}=8, \mathrm{r}=4\)
3 \(\mathrm{n}=7, \mathrm{r}=5\)
4 \(\mathrm{r}=6, \mathrm{n}=16\)
Explanation:
A Given that, \({ }^n C_{r-1}=36\) \({ }^n C_r=84\) \({ }^n C_{r+1}=126\) Divide equation (ii) by equation (i), we get \(\frac{{ }^n C_r}{{ }^n C_r-1}=\frac{84}{36}\) \(\Rightarrow \quad \frac{\mathrm{n}-\mathrm{r}+1}{\mathrm{r}}=\frac{7}{3}\) \(\Rightarrow \quad 3 \mathrm{n}-3 \mathrm{r}+3=7 \mathrm{r}\) \(\Rightarrow \quad 3 \mathrm{n}-10 \mathrm{r}+3=0\) Divide equation (iii) by equation (ii), we get :- \(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\) \(\Rightarrow \frac{n-r}{r+1}=\frac{3}{2}\) \(\Rightarrow \quad 2 n-2 r=3 r+3\) \(\Rightarrow \quad 2 n-5 r-3=0\) Solving equation (iv) and (v), we get \(\mathrm{n}=9, \mathrm{r}=3\)
SRM JEEE-2009
Permutation and Combination
118909
If \({ }^n P_r=720{ }^n C_r\), then the value of \(r\) is
1 6
2 5
3 4
4 7
Explanation:
A Given that, \({ }^{{ }^n P_{\mathrm{r}}}=720{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) \(\Rightarrow \quad \frac{n !}{(n-r) !}=720 \times \frac{n !}{(n-r) ! r !}\) \(\mathrm{r} !=720\) \(\Rightarrow \quad \mathrm{r} !=6\) ! \(\Rightarrow \quad \mathrm{r}=6\)
SRM JEEE-2011
Permutation and Combination
118910
The number of positive integral solution of abc \(=30\) is
1 30
2 27
3 8
4 None of these
Explanation:
B Given the positive integral solution, \(\mathrm{abc}=30\) \(\therefore \quad 30=2 \times 3 \times 5\) \(=5 \times 6 \times 1\) \(=15 \times 2 \times 1\) \(=30 \times 1 \times 1\) Here the possibility of 30 is three. It can be as a, b, c and the possibility of 1 can be both of the digits at time. So, this is written as- \(\frac{3 !}{2 !}\) So, number of solution is- \(=3 !+3 !+3 !+3 !+\frac{3 !}{2 !}\) \(=(4 \times 3 !)+\frac{3 !}{2 !}\) \(=(4 \times 6)+3\) \(=24+3\) \(=27\)