119253
The sum of digits in the unit's place of all the numbers formed with the help of \(3,4,5,6\) taken all at a time is
1 432
2 36
3 18
4 108
Explanation:
D Here number of numbers formed when 3 is at units place is \(3 \times 2 \times 1=6\) Similarly, no. of numbers formed when 4,5 and 6 is at units place is 6 each. So, required sum \(=6(3+4+5+6)\) \(=6 \times 18\) \(=108\)
SRM JEEE-2009
Permutation and Combination
119254
In a certain test there are \(n\) questions. In this test \(2^{\mathrm{n}-\mathrm{k}}\) students gave wrong answers to \(\mathrm{k}\) questions, where \(k=1,2,3, \ldots . . . . ., n\). If the total number of wrong answers given is 2047 , then \(n\) is equal to
1 10
2 11
3 12
4 13
Explanation:
B We have, Number of student given wrong answer \(=2^{\mathrm{n}-\mathrm{k}}\) Where \(\mathrm{k}=1,2,3 \ldots\). Total number of wrong answer given \(=2047\) Total no. of wrong answers \(=2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+\ldots .+2^2+2^1+2^0\) \(2047=\frac{1\left(2^{\mathrm{n}}-1\right)}{2-1}\) \(2^{\mathrm{n}}-1=2047\) \(2^{\mathrm{n}}=2048\) \(2^{\mathrm{n}}=2^{11}\) \(\mathrm{n}=11\)
SRM JEEE-2008
Permutation and Combination
119255
There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelopes is
1 8
2 16
3 24
4 9
Explanation:
D Given that, \(\text { Number of letter }=4\) \(\text { Number of envelops }=4\) \(\text { It is problem of derangement. }\) \(\text { Thus, we use the formula }\) \(\mathrm{D}_{\mathrm{n}}=\mathrm{n} !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \frac{(-1)^{\mathrm{n}}}{\mathrm{n} !}\right)\) \(\text { Therefore, required number of ways }\) \(\mathrm{D}_4=4 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]\) \(\quad=24\left[1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right]\) \(\quad=24\left[\frac{12-4+1}{24}\right]\) \(\quad=8+1\) \(=9\)
119253
The sum of digits in the unit's place of all the numbers formed with the help of \(3,4,5,6\) taken all at a time is
1 432
2 36
3 18
4 108
Explanation:
D Here number of numbers formed when 3 is at units place is \(3 \times 2 \times 1=6\) Similarly, no. of numbers formed when 4,5 and 6 is at units place is 6 each. So, required sum \(=6(3+4+5+6)\) \(=6 \times 18\) \(=108\)
SRM JEEE-2009
Permutation and Combination
119254
In a certain test there are \(n\) questions. In this test \(2^{\mathrm{n}-\mathrm{k}}\) students gave wrong answers to \(\mathrm{k}\) questions, where \(k=1,2,3, \ldots . . . . ., n\). If the total number of wrong answers given is 2047 , then \(n\) is equal to
1 10
2 11
3 12
4 13
Explanation:
B We have, Number of student given wrong answer \(=2^{\mathrm{n}-\mathrm{k}}\) Where \(\mathrm{k}=1,2,3 \ldots\). Total number of wrong answer given \(=2047\) Total no. of wrong answers \(=2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+\ldots .+2^2+2^1+2^0\) \(2047=\frac{1\left(2^{\mathrm{n}}-1\right)}{2-1}\) \(2^{\mathrm{n}}-1=2047\) \(2^{\mathrm{n}}=2048\) \(2^{\mathrm{n}}=2^{11}\) \(\mathrm{n}=11\)
SRM JEEE-2008
Permutation and Combination
119255
There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelopes is
1 8
2 16
3 24
4 9
Explanation:
D Given that, \(\text { Number of letter }=4\) \(\text { Number of envelops }=4\) \(\text { It is problem of derangement. }\) \(\text { Thus, we use the formula }\) \(\mathrm{D}_{\mathrm{n}}=\mathrm{n} !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \frac{(-1)^{\mathrm{n}}}{\mathrm{n} !}\right)\) \(\text { Therefore, required number of ways }\) \(\mathrm{D}_4=4 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]\) \(\quad=24\left[1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right]\) \(\quad=24\left[\frac{12-4+1}{24}\right]\) \(\quad=8+1\) \(=9\)
119253
The sum of digits in the unit's place of all the numbers formed with the help of \(3,4,5,6\) taken all at a time is
1 432
2 36
3 18
4 108
Explanation:
D Here number of numbers formed when 3 is at units place is \(3 \times 2 \times 1=6\) Similarly, no. of numbers formed when 4,5 and 6 is at units place is 6 each. So, required sum \(=6(3+4+5+6)\) \(=6 \times 18\) \(=108\)
SRM JEEE-2009
Permutation and Combination
119254
In a certain test there are \(n\) questions. In this test \(2^{\mathrm{n}-\mathrm{k}}\) students gave wrong answers to \(\mathrm{k}\) questions, where \(k=1,2,3, \ldots . . . . ., n\). If the total number of wrong answers given is 2047 , then \(n\) is equal to
1 10
2 11
3 12
4 13
Explanation:
B We have, Number of student given wrong answer \(=2^{\mathrm{n}-\mathrm{k}}\) Where \(\mathrm{k}=1,2,3 \ldots\). Total number of wrong answer given \(=2047\) Total no. of wrong answers \(=2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+\ldots .+2^2+2^1+2^0\) \(2047=\frac{1\left(2^{\mathrm{n}}-1\right)}{2-1}\) \(2^{\mathrm{n}}-1=2047\) \(2^{\mathrm{n}}=2048\) \(2^{\mathrm{n}}=2^{11}\) \(\mathrm{n}=11\)
SRM JEEE-2008
Permutation and Combination
119255
There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelopes is
1 8
2 16
3 24
4 9
Explanation:
D Given that, \(\text { Number of letter }=4\) \(\text { Number of envelops }=4\) \(\text { It is problem of derangement. }\) \(\text { Thus, we use the formula }\) \(\mathrm{D}_{\mathrm{n}}=\mathrm{n} !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \frac{(-1)^{\mathrm{n}}}{\mathrm{n} !}\right)\) \(\text { Therefore, required number of ways }\) \(\mathrm{D}_4=4 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]\) \(\quad=24\left[1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right]\) \(\quad=24\left[\frac{12-4+1}{24}\right]\) \(\quad=8+1\) \(=9\)
119253
The sum of digits in the unit's place of all the numbers formed with the help of \(3,4,5,6\) taken all at a time is
1 432
2 36
3 18
4 108
Explanation:
D Here number of numbers formed when 3 is at units place is \(3 \times 2 \times 1=6\) Similarly, no. of numbers formed when 4,5 and 6 is at units place is 6 each. So, required sum \(=6(3+4+5+6)\) \(=6 \times 18\) \(=108\)
SRM JEEE-2009
Permutation and Combination
119254
In a certain test there are \(n\) questions. In this test \(2^{\mathrm{n}-\mathrm{k}}\) students gave wrong answers to \(\mathrm{k}\) questions, where \(k=1,2,3, \ldots . . . . ., n\). If the total number of wrong answers given is 2047 , then \(n\) is equal to
1 10
2 11
3 12
4 13
Explanation:
B We have, Number of student given wrong answer \(=2^{\mathrm{n}-\mathrm{k}}\) Where \(\mathrm{k}=1,2,3 \ldots\). Total number of wrong answer given \(=2047\) Total no. of wrong answers \(=2^{\mathrm{n}-1}+2^{\mathrm{n}-2}+\ldots .+2^2+2^1+2^0\) \(2047=\frac{1\left(2^{\mathrm{n}}-1\right)}{2-1}\) \(2^{\mathrm{n}}-1=2047\) \(2^{\mathrm{n}}=2048\) \(2^{\mathrm{n}}=2^{11}\) \(\mathrm{n}=11\)
SRM JEEE-2008
Permutation and Combination
119255
There are 4 letters and 4 directed envelopes. The number of ways in which all the letters can be put in wrong envelopes is
1 8
2 16
3 24
4 9
Explanation:
D Given that, \(\text { Number of letter }=4\) \(\text { Number of envelops }=4\) \(\text { It is problem of derangement. }\) \(\text { Thus, we use the formula }\) \(\mathrm{D}_{\mathrm{n}}=\mathrm{n} !\left(1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\ldots \frac{(-1)^{\mathrm{n}}}{\mathrm{n} !}\right)\) \(\text { Therefore, required number of ways }\) \(\mathrm{D}_4=4 !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}\right]\) \(\quad=24\left[1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}\right]\) \(\quad=24\left[\frac{12-4+1}{24}\right]\) \(\quad=8+1\) \(=9\)
