C We know that, \(7^1=7 \Rightarrow\) last digit \(=7\) \(7^2=49 \Rightarrow\) last digit \(=9\) \(7^3=343 \Rightarrow\) last digit \(=3\) \(7^4=2401 \Rightarrow\) last digit \(=1\) \(7^5=16807 \Rightarrow\) last digit \(=7\) and so on. Last digit of \(7^{300}=\frac{300}{4}\) Thus, we get that 300 is divisible by 4 and the unit digit is 1 Thus, we can conclude that the unit digit of \(7^{300}\) is 1 .
Karnataka CET-2004
Permutation and Combination
118964
How many number of 6 digits can be formed from the digits of the number 112233 ?
1 30
2 60
3 90
4 120
Explanation:
C Given \( \text{number }=112233\) \(\text { Repeated digit }=1,2,3\) \(\text { Total number of digit }=6\) \(\text { So, the total number of digits that can be formed by }\) \(112233 =\frac{6 !}{2 ! \times 2 ! \times 2 !}\) \(=\frac{6 \times 5 \times 4 \times 3}{4}\) \(=90\)
Karnataka CET-2004
Permutation and Combination
118965
A student has to answer \(\mathbf{1 0}\) questions, choosing at least 4 from each of the parts \(A\) and \(B\). If there are 6 questions in part \(A\) and 7 in part \(B\), then the number of ways can the student choose 10 questions is
1 256
2 352
3 266
4 426
Explanation:
C Here, total number of questions \(=10\) \(\text { Total number of questions in part } \mathrm{A}=6\) \(\text { Total number of question in part } \mathrm{B}=7\) \(\text { Total no of ways can the student choose } 10 \text { questions. }\) \(=\left({ }^6 \mathrm{C}_4 \times{ }^7 \mathrm{C}_6\right)+\left({ }^6 \mathrm{C}_5 \times{ }^7 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_6 \times{ }^7 \mathrm{C}_4\right)\) \(\text { We know that, }{ }^n \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\) \(=\left(\frac{6 !}{4 ! \times 2 !} \times \frac{7 !}{6 ! 1 !}\right)+\left(\frac{6 !}{5 ! \times 1 !} \times \frac{7 !}{5 ! 2 !}\right)+\left(\frac{6 !}{6 ! \times 0 !} \times \frac{7 !}{4 ! \times 3 !}\right)\) \(=(15 \times 7)+(6 \times 21)+(1 \times 35)\) \(=105+126+35\) \(=266\)
Karnataka CET-2021
Permutation and Combination
118966
The number of ways in which ten candidates \(A_1, A_2 \ldots . . . A_{10}\) can be ranked such that \(A_1\) is always above \(A_{10}\) is
1 5 !
2 \(2(5 !)\)
3 10 !
4 \(\frac{1}{2}(10 !)\)
Explanation:
D Without any restriction the 10 persons can be ranked among themselves is 10 ! Ways. There are two cases; One in which \(\mathrm{A}_1\) is above \(\mathrm{A}_{10}\) and other is \(\mathrm{A}_1\) is below \(\mathrm{A}_{10}\) \(\therefore\) Number of ways when \(\mathrm{A}_1\) is always above \(\mathrm{A}_{10}=\frac{1}{2}(10 !)\) ways
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Permutation and Combination
118963
The last digit in \(7^{300}\) is :
1 7
2 9
3 1
4 3
Explanation:
C We know that, \(7^1=7 \Rightarrow\) last digit \(=7\) \(7^2=49 \Rightarrow\) last digit \(=9\) \(7^3=343 \Rightarrow\) last digit \(=3\) \(7^4=2401 \Rightarrow\) last digit \(=1\) \(7^5=16807 \Rightarrow\) last digit \(=7\) and so on. Last digit of \(7^{300}=\frac{300}{4}\) Thus, we get that 300 is divisible by 4 and the unit digit is 1 Thus, we can conclude that the unit digit of \(7^{300}\) is 1 .
Karnataka CET-2004
Permutation and Combination
118964
How many number of 6 digits can be formed from the digits of the number 112233 ?
1 30
2 60
3 90
4 120
Explanation:
C Given \( \text{number }=112233\) \(\text { Repeated digit }=1,2,3\) \(\text { Total number of digit }=6\) \(\text { So, the total number of digits that can be formed by }\) \(112233 =\frac{6 !}{2 ! \times 2 ! \times 2 !}\) \(=\frac{6 \times 5 \times 4 \times 3}{4}\) \(=90\)
Karnataka CET-2004
Permutation and Combination
118965
A student has to answer \(\mathbf{1 0}\) questions, choosing at least 4 from each of the parts \(A\) and \(B\). If there are 6 questions in part \(A\) and 7 in part \(B\), then the number of ways can the student choose 10 questions is
1 256
2 352
3 266
4 426
Explanation:
C Here, total number of questions \(=10\) \(\text { Total number of questions in part } \mathrm{A}=6\) \(\text { Total number of question in part } \mathrm{B}=7\) \(\text { Total no of ways can the student choose } 10 \text { questions. }\) \(=\left({ }^6 \mathrm{C}_4 \times{ }^7 \mathrm{C}_6\right)+\left({ }^6 \mathrm{C}_5 \times{ }^7 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_6 \times{ }^7 \mathrm{C}_4\right)\) \(\text { We know that, }{ }^n \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\) \(=\left(\frac{6 !}{4 ! \times 2 !} \times \frac{7 !}{6 ! 1 !}\right)+\left(\frac{6 !}{5 ! \times 1 !} \times \frac{7 !}{5 ! 2 !}\right)+\left(\frac{6 !}{6 ! \times 0 !} \times \frac{7 !}{4 ! \times 3 !}\right)\) \(=(15 \times 7)+(6 \times 21)+(1 \times 35)\) \(=105+126+35\) \(=266\)
Karnataka CET-2021
Permutation and Combination
118966
The number of ways in which ten candidates \(A_1, A_2 \ldots . . . A_{10}\) can be ranked such that \(A_1\) is always above \(A_{10}\) is
1 5 !
2 \(2(5 !)\)
3 10 !
4 \(\frac{1}{2}(10 !)\)
Explanation:
D Without any restriction the 10 persons can be ranked among themselves is 10 ! Ways. There are two cases; One in which \(\mathrm{A}_1\) is above \(\mathrm{A}_{10}\) and other is \(\mathrm{A}_1\) is below \(\mathrm{A}_{10}\) \(\therefore\) Number of ways when \(\mathrm{A}_1\) is always above \(\mathrm{A}_{10}=\frac{1}{2}(10 !)\) ways
C We know that, \(7^1=7 \Rightarrow\) last digit \(=7\) \(7^2=49 \Rightarrow\) last digit \(=9\) \(7^3=343 \Rightarrow\) last digit \(=3\) \(7^4=2401 \Rightarrow\) last digit \(=1\) \(7^5=16807 \Rightarrow\) last digit \(=7\) and so on. Last digit of \(7^{300}=\frac{300}{4}\) Thus, we get that 300 is divisible by 4 and the unit digit is 1 Thus, we can conclude that the unit digit of \(7^{300}\) is 1 .
Karnataka CET-2004
Permutation and Combination
118964
How many number of 6 digits can be formed from the digits of the number 112233 ?
1 30
2 60
3 90
4 120
Explanation:
C Given \( \text{number }=112233\) \(\text { Repeated digit }=1,2,3\) \(\text { Total number of digit }=6\) \(\text { So, the total number of digits that can be formed by }\) \(112233 =\frac{6 !}{2 ! \times 2 ! \times 2 !}\) \(=\frac{6 \times 5 \times 4 \times 3}{4}\) \(=90\)
Karnataka CET-2004
Permutation and Combination
118965
A student has to answer \(\mathbf{1 0}\) questions, choosing at least 4 from each of the parts \(A\) and \(B\). If there are 6 questions in part \(A\) and 7 in part \(B\), then the number of ways can the student choose 10 questions is
1 256
2 352
3 266
4 426
Explanation:
C Here, total number of questions \(=10\) \(\text { Total number of questions in part } \mathrm{A}=6\) \(\text { Total number of question in part } \mathrm{B}=7\) \(\text { Total no of ways can the student choose } 10 \text { questions. }\) \(=\left({ }^6 \mathrm{C}_4 \times{ }^7 \mathrm{C}_6\right)+\left({ }^6 \mathrm{C}_5 \times{ }^7 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_6 \times{ }^7 \mathrm{C}_4\right)\) \(\text { We know that, }{ }^n \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\) \(=\left(\frac{6 !}{4 ! \times 2 !} \times \frac{7 !}{6 ! 1 !}\right)+\left(\frac{6 !}{5 ! \times 1 !} \times \frac{7 !}{5 ! 2 !}\right)+\left(\frac{6 !}{6 ! \times 0 !} \times \frac{7 !}{4 ! \times 3 !}\right)\) \(=(15 \times 7)+(6 \times 21)+(1 \times 35)\) \(=105+126+35\) \(=266\)
Karnataka CET-2021
Permutation and Combination
118966
The number of ways in which ten candidates \(A_1, A_2 \ldots . . . A_{10}\) can be ranked such that \(A_1\) is always above \(A_{10}\) is
1 5 !
2 \(2(5 !)\)
3 10 !
4 \(\frac{1}{2}(10 !)\)
Explanation:
D Without any restriction the 10 persons can be ranked among themselves is 10 ! Ways. There are two cases; One in which \(\mathrm{A}_1\) is above \(\mathrm{A}_{10}\) and other is \(\mathrm{A}_1\) is below \(\mathrm{A}_{10}\) \(\therefore\) Number of ways when \(\mathrm{A}_1\) is always above \(\mathrm{A}_{10}=\frac{1}{2}(10 !)\) ways
C We know that, \(7^1=7 \Rightarrow\) last digit \(=7\) \(7^2=49 \Rightarrow\) last digit \(=9\) \(7^3=343 \Rightarrow\) last digit \(=3\) \(7^4=2401 \Rightarrow\) last digit \(=1\) \(7^5=16807 \Rightarrow\) last digit \(=7\) and so on. Last digit of \(7^{300}=\frac{300}{4}\) Thus, we get that 300 is divisible by 4 and the unit digit is 1 Thus, we can conclude that the unit digit of \(7^{300}\) is 1 .
Karnataka CET-2004
Permutation and Combination
118964
How many number of 6 digits can be formed from the digits of the number 112233 ?
1 30
2 60
3 90
4 120
Explanation:
C Given \( \text{number }=112233\) \(\text { Repeated digit }=1,2,3\) \(\text { Total number of digit }=6\) \(\text { So, the total number of digits that can be formed by }\) \(112233 =\frac{6 !}{2 ! \times 2 ! \times 2 !}\) \(=\frac{6 \times 5 \times 4 \times 3}{4}\) \(=90\)
Karnataka CET-2004
Permutation and Combination
118965
A student has to answer \(\mathbf{1 0}\) questions, choosing at least 4 from each of the parts \(A\) and \(B\). If there are 6 questions in part \(A\) and 7 in part \(B\), then the number of ways can the student choose 10 questions is
1 256
2 352
3 266
4 426
Explanation:
C Here, total number of questions \(=10\) \(\text { Total number of questions in part } \mathrm{A}=6\) \(\text { Total number of question in part } \mathrm{B}=7\) \(\text { Total no of ways can the student choose } 10 \text { questions. }\) \(=\left({ }^6 \mathrm{C}_4 \times{ }^7 \mathrm{C}_6\right)+\left({ }^6 \mathrm{C}_5 \times{ }^7 \mathrm{C}_5\right)+\left({ }^6 \mathrm{C}_6 \times{ }^7 \mathrm{C}_4\right)\) \(\text { We know that, }{ }^n \mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n} !}{\mathrm{r} !(\mathrm{n}-\mathrm{r}) !}\) \(=\left(\frac{6 !}{4 ! \times 2 !} \times \frac{7 !}{6 ! 1 !}\right)+\left(\frac{6 !}{5 ! \times 1 !} \times \frac{7 !}{5 ! 2 !}\right)+\left(\frac{6 !}{6 ! \times 0 !} \times \frac{7 !}{4 ! \times 3 !}\right)\) \(=(15 \times 7)+(6 \times 21)+(1 \times 35)\) \(=105+126+35\) \(=266\)
Karnataka CET-2021
Permutation and Combination
118966
The number of ways in which ten candidates \(A_1, A_2 \ldots . . . A_{10}\) can be ranked such that \(A_1\) is always above \(A_{10}\) is
1 5 !
2 \(2(5 !)\)
3 10 !
4 \(\frac{1}{2}(10 !)\)
Explanation:
D Without any restriction the 10 persons can be ranked among themselves is 10 ! Ways. There are two cases; One in which \(\mathrm{A}_1\) is above \(\mathrm{A}_{10}\) and other is \(\mathrm{A}_1\) is below \(\mathrm{A}_{10}\) \(\therefore\) Number of ways when \(\mathrm{A}_1\) is always above \(\mathrm{A}_{10}=\frac{1}{2}(10 !)\) ways