NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
119021
Four dice are rolled. The number of possible outcomes in which at least one dice shows 2 is
1 625
2 671
3 1023
4 1296
Explanation:
B When 4 dice are rolled then total number of outcome \(=6^4\) Number of outcomes when no two dice show \(=5^4\) Favorable case \(=\) Total cases - unfavorable cases \(=6^4-5^4\) \(=1296-625\) \(=671\)
AMU-2009
Permutation and Combination
119022
How many committees of five with a chairperson can be selected from 12 persons?
1 330
2 3630
3 3960
4 none of the above
Explanation:
C Number of person to be selected \(=5\) out of 5 , Now selecting a chair person \(={ }^{12} \mathrm{C}_1\) Number of way selecting other 4 person out of remaining 11 person \(={ }^{11} \mathrm{C}_4\) Total number of ways - \(={ }^{12} \mathrm{C}_1 \times{ }^{11} \mathrm{C}_4\) \(=12 \times \frac{\lfloor 11}{\lfloor! 4}\) \(12 \times \frac{11 \times 10 \times 9 \times 8 \times 7 !}{7 ! 4 !}=12 \times 330=3960\)
AMU-2006
Permutation and Combination
119023
There are \(n\) students of \(B\). Tech. \(-1^{\text {st }}\) year and n students of B. Tech. \(-2^{\text {nd }}\) years. The number of ways of arranging all the \(2 n\) students in a row so that neighbouring students are of different classes is
1 \(2(n !)^2\)
2 \((2 n) ! /(n !)^2\)
3 \(\frac{1}{2}(2 n)\) !
4 \((2 n) ! /(n !)\)
Explanation:
A Given that, \(\mathrm{n}\) student of \(1^{\mathrm{st}}\) year and \(\mathrm{n}\) student of \(2^{\text {nd }}\) year. Since, \(\mathrm{n}\) student from \(1^{\text {st }}\) year arranging in \(\mathrm{n}\) ! ways and \(\mathrm{n}\) student from \(2^{\text {nd }}\) year arranging in \(\mathrm{n}\) ! ways. The total number of arrangement \(\rightarrow \mathrm{n} ! \times \mathrm{n}\) ! Now, there are two possible ways of arrangement \(\therefore\) Total number of ways \(=2(n !)^2\)
AMU-2006
Permutation and Combination
119024
The number of ways that 8 beads of different colours be strung as a necklace is
1 2520
2 2880
3 4320
4 5040
Explanation:
A Number of ways that 8 different beads can be arranged \(=\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(\text { Number of ways that } 8 \text { different beads can }\) \(\text { be arranged } =\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) \(=2520\)
119021
Four dice are rolled. The number of possible outcomes in which at least one dice shows 2 is
1 625
2 671
3 1023
4 1296
Explanation:
B When 4 dice are rolled then total number of outcome \(=6^4\) Number of outcomes when no two dice show \(=5^4\) Favorable case \(=\) Total cases - unfavorable cases \(=6^4-5^4\) \(=1296-625\) \(=671\)
AMU-2009
Permutation and Combination
119022
How many committees of five with a chairperson can be selected from 12 persons?
1 330
2 3630
3 3960
4 none of the above
Explanation:
C Number of person to be selected \(=5\) out of 5 , Now selecting a chair person \(={ }^{12} \mathrm{C}_1\) Number of way selecting other 4 person out of remaining 11 person \(={ }^{11} \mathrm{C}_4\) Total number of ways - \(={ }^{12} \mathrm{C}_1 \times{ }^{11} \mathrm{C}_4\) \(=12 \times \frac{\lfloor 11}{\lfloor! 4}\) \(12 \times \frac{11 \times 10 \times 9 \times 8 \times 7 !}{7 ! 4 !}=12 \times 330=3960\)
AMU-2006
Permutation and Combination
119023
There are \(n\) students of \(B\). Tech. \(-1^{\text {st }}\) year and n students of B. Tech. \(-2^{\text {nd }}\) years. The number of ways of arranging all the \(2 n\) students in a row so that neighbouring students are of different classes is
1 \(2(n !)^2\)
2 \((2 n) ! /(n !)^2\)
3 \(\frac{1}{2}(2 n)\) !
4 \((2 n) ! /(n !)\)
Explanation:
A Given that, \(\mathrm{n}\) student of \(1^{\mathrm{st}}\) year and \(\mathrm{n}\) student of \(2^{\text {nd }}\) year. Since, \(\mathrm{n}\) student from \(1^{\text {st }}\) year arranging in \(\mathrm{n}\) ! ways and \(\mathrm{n}\) student from \(2^{\text {nd }}\) year arranging in \(\mathrm{n}\) ! ways. The total number of arrangement \(\rightarrow \mathrm{n} ! \times \mathrm{n}\) ! Now, there are two possible ways of arrangement \(\therefore\) Total number of ways \(=2(n !)^2\)
AMU-2006
Permutation and Combination
119024
The number of ways that 8 beads of different colours be strung as a necklace is
1 2520
2 2880
3 4320
4 5040
Explanation:
A Number of ways that 8 different beads can be arranged \(=\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(\text { Number of ways that } 8 \text { different beads can }\) \(\text { be arranged } =\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) \(=2520\)
119021
Four dice are rolled. The number of possible outcomes in which at least one dice shows 2 is
1 625
2 671
3 1023
4 1296
Explanation:
B When 4 dice are rolled then total number of outcome \(=6^4\) Number of outcomes when no two dice show \(=5^4\) Favorable case \(=\) Total cases - unfavorable cases \(=6^4-5^4\) \(=1296-625\) \(=671\)
AMU-2009
Permutation and Combination
119022
How many committees of five with a chairperson can be selected from 12 persons?
1 330
2 3630
3 3960
4 none of the above
Explanation:
C Number of person to be selected \(=5\) out of 5 , Now selecting a chair person \(={ }^{12} \mathrm{C}_1\) Number of way selecting other 4 person out of remaining 11 person \(={ }^{11} \mathrm{C}_4\) Total number of ways - \(={ }^{12} \mathrm{C}_1 \times{ }^{11} \mathrm{C}_4\) \(=12 \times \frac{\lfloor 11}{\lfloor! 4}\) \(12 \times \frac{11 \times 10 \times 9 \times 8 \times 7 !}{7 ! 4 !}=12 \times 330=3960\)
AMU-2006
Permutation and Combination
119023
There are \(n\) students of \(B\). Tech. \(-1^{\text {st }}\) year and n students of B. Tech. \(-2^{\text {nd }}\) years. The number of ways of arranging all the \(2 n\) students in a row so that neighbouring students are of different classes is
1 \(2(n !)^2\)
2 \((2 n) ! /(n !)^2\)
3 \(\frac{1}{2}(2 n)\) !
4 \((2 n) ! /(n !)\)
Explanation:
A Given that, \(\mathrm{n}\) student of \(1^{\mathrm{st}}\) year and \(\mathrm{n}\) student of \(2^{\text {nd }}\) year. Since, \(\mathrm{n}\) student from \(1^{\text {st }}\) year arranging in \(\mathrm{n}\) ! ways and \(\mathrm{n}\) student from \(2^{\text {nd }}\) year arranging in \(\mathrm{n}\) ! ways. The total number of arrangement \(\rightarrow \mathrm{n} ! \times \mathrm{n}\) ! Now, there are two possible ways of arrangement \(\therefore\) Total number of ways \(=2(n !)^2\)
AMU-2006
Permutation and Combination
119024
The number of ways that 8 beads of different colours be strung as a necklace is
1 2520
2 2880
3 4320
4 5040
Explanation:
A Number of ways that 8 different beads can be arranged \(=\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(\text { Number of ways that } 8 \text { different beads can }\) \(\text { be arranged } =\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) \(=2520\)
119021
Four dice are rolled. The number of possible outcomes in which at least one dice shows 2 is
1 625
2 671
3 1023
4 1296
Explanation:
B When 4 dice are rolled then total number of outcome \(=6^4\) Number of outcomes when no two dice show \(=5^4\) Favorable case \(=\) Total cases - unfavorable cases \(=6^4-5^4\) \(=1296-625\) \(=671\)
AMU-2009
Permutation and Combination
119022
How many committees of five with a chairperson can be selected from 12 persons?
1 330
2 3630
3 3960
4 none of the above
Explanation:
C Number of person to be selected \(=5\) out of 5 , Now selecting a chair person \(={ }^{12} \mathrm{C}_1\) Number of way selecting other 4 person out of remaining 11 person \(={ }^{11} \mathrm{C}_4\) Total number of ways - \(={ }^{12} \mathrm{C}_1 \times{ }^{11} \mathrm{C}_4\) \(=12 \times \frac{\lfloor 11}{\lfloor! 4}\) \(12 \times \frac{11 \times 10 \times 9 \times 8 \times 7 !}{7 ! 4 !}=12 \times 330=3960\)
AMU-2006
Permutation and Combination
119023
There are \(n\) students of \(B\). Tech. \(-1^{\text {st }}\) year and n students of B. Tech. \(-2^{\text {nd }}\) years. The number of ways of arranging all the \(2 n\) students in a row so that neighbouring students are of different classes is
1 \(2(n !)^2\)
2 \((2 n) ! /(n !)^2\)
3 \(\frac{1}{2}(2 n)\) !
4 \((2 n) ! /(n !)\)
Explanation:
A Given that, \(\mathrm{n}\) student of \(1^{\mathrm{st}}\) year and \(\mathrm{n}\) student of \(2^{\text {nd }}\) year. Since, \(\mathrm{n}\) student from \(1^{\text {st }}\) year arranging in \(\mathrm{n}\) ! ways and \(\mathrm{n}\) student from \(2^{\text {nd }}\) year arranging in \(\mathrm{n}\) ! ways. The total number of arrangement \(\rightarrow \mathrm{n} ! \times \mathrm{n}\) ! Now, there are two possible ways of arrangement \(\therefore\) Total number of ways \(=2(n !)^2\)
AMU-2006
Permutation and Combination
119024
The number of ways that 8 beads of different colours be strung as a necklace is
1 2520
2 2880
3 4320
4 5040
Explanation:
A Number of ways that 8 different beads can be arranged \(=\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(\text { Number of ways that } 8 \text { different beads can }\) \(\text { be arranged } =\frac{(8-1) !}{2}=\frac{7 !}{2}\) \(=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{2}\) \(=2520\)
