119277
If \({ }^n p_4=20 \times{ }^n p_2\). Then the value of \(n\) is
1 18
2 13
3 7
4 4
Explanation:
C We have \({ }^n p_4=20 \times{ }^n p_2\) \(\frac{n !}{(n-4) !}=20 \times \frac{n !}{(n-2) !}\) \(\frac{1}{(n-4) !}=20 \times \frac{1}{(n-2)(n-3)(n-4) !}\) \((n-2)(n-3)=20\) \(n^2-3 n-2 n+6=20\) \(n^2-5 n-14=0\) \((n-7)(n+2)=0\) \(n=7\)
AMU-2017
Permutation and Combination
119279
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is
1 20
2 36
3 30
4 720
Explanation:
B Given that, \(\text { Total word in MOBILE }=6 !\) \(\text { Number of consonants }=\mathrm{M}, \mathrm{B}, \mathrm{L}=3\) \(\text { Number of vowel = O,I,E = 3 }\) At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words \(=3 ! \times 3 !=6 \times 6=36\)
APEAPCET- 23.08.2021
Permutation and Combination
119280
If ' \(n\) ' is a positive integer, then \(2.4^{2 n+1}+3^{3 n+1}\) is divisible by
1 2
2 9
3 11
4 27
Explanation:
C : Given that, \(\mathrm{n}\) is a positive integer If \(\mathrm{n}=1\) \(2.4^{2(1)+1}+3^{3(1)+1}\) \(=2.4^3+3^4\) \(=2.64+3^4\) \(=2.64+81=128+81=209\) Hence, 209 is the divisible by 11 .
119277
If \({ }^n p_4=20 \times{ }^n p_2\). Then the value of \(n\) is
1 18
2 13
3 7
4 4
Explanation:
C We have \({ }^n p_4=20 \times{ }^n p_2\) \(\frac{n !}{(n-4) !}=20 \times \frac{n !}{(n-2) !}\) \(\frac{1}{(n-4) !}=20 \times \frac{1}{(n-2)(n-3)(n-4) !}\) \((n-2)(n-3)=20\) \(n^2-3 n-2 n+6=20\) \(n^2-5 n-14=0\) \((n-7)(n+2)=0\) \(n=7\)
AMU-2017
Permutation and Combination
119279
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is
1 20
2 36
3 30
4 720
Explanation:
B Given that, \(\text { Total word in MOBILE }=6 !\) \(\text { Number of consonants }=\mathrm{M}, \mathrm{B}, \mathrm{L}=3\) \(\text { Number of vowel = O,I,E = 3 }\) At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words \(=3 ! \times 3 !=6 \times 6=36\)
APEAPCET- 23.08.2021
Permutation and Combination
119280
If ' \(n\) ' is a positive integer, then \(2.4^{2 n+1}+3^{3 n+1}\) is divisible by
1 2
2 9
3 11
4 27
Explanation:
C : Given that, \(\mathrm{n}\) is a positive integer If \(\mathrm{n}=1\) \(2.4^{2(1)+1}+3^{3(1)+1}\) \(=2.4^3+3^4\) \(=2.64+3^4\) \(=2.64+81=128+81=209\) Hence, 209 is the divisible by 11 .
119277
If \({ }^n p_4=20 \times{ }^n p_2\). Then the value of \(n\) is
1 18
2 13
3 7
4 4
Explanation:
C We have \({ }^n p_4=20 \times{ }^n p_2\) \(\frac{n !}{(n-4) !}=20 \times \frac{n !}{(n-2) !}\) \(\frac{1}{(n-4) !}=20 \times \frac{1}{(n-2)(n-3)(n-4) !}\) \((n-2)(n-3)=20\) \(n^2-3 n-2 n+6=20\) \(n^2-5 n-14=0\) \((n-7)(n+2)=0\) \(n=7\)
AMU-2017
Permutation and Combination
119279
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is
1 20
2 36
3 30
4 720
Explanation:
B Given that, \(\text { Total word in MOBILE }=6 !\) \(\text { Number of consonants }=\mathrm{M}, \mathrm{B}, \mathrm{L}=3\) \(\text { Number of vowel = O,I,E = 3 }\) At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words \(=3 ! \times 3 !=6 \times 6=36\)
APEAPCET- 23.08.2021
Permutation and Combination
119280
If ' \(n\) ' is a positive integer, then \(2.4^{2 n+1}+3^{3 n+1}\) is divisible by
1 2
2 9
3 11
4 27
Explanation:
C : Given that, \(\mathrm{n}\) is a positive integer If \(\mathrm{n}=1\) \(2.4^{2(1)+1}+3^{3(1)+1}\) \(=2.4^3+3^4\) \(=2.64+3^4\) \(=2.64+81=128+81=209\) Hence, 209 is the divisible by 11 .
119277
If \({ }^n p_4=20 \times{ }^n p_2\). Then the value of \(n\) is
1 18
2 13
3 7
4 4
Explanation:
C We have \({ }^n p_4=20 \times{ }^n p_2\) \(\frac{n !}{(n-4) !}=20 \times \frac{n !}{(n-2) !}\) \(\frac{1}{(n-4) !}=20 \times \frac{1}{(n-2)(n-3)(n-4) !}\) \((n-2)(n-3)=20\) \(n^2-3 n-2 n+6=20\) \(n^2-5 n-14=0\) \((n-7)(n+2)=0\) \(n=7\)
AMU-2017
Permutation and Combination
119279
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is
1 20
2 36
3 30
4 720
Explanation:
B Given that, \(\text { Total word in MOBILE }=6 !\) \(\text { Number of consonants }=\mathrm{M}, \mathrm{B}, \mathrm{L}=3\) \(\text { Number of vowel = O,I,E = 3 }\) At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words \(=3 ! \times 3 !=6 \times 6=36\)
APEAPCET- 23.08.2021
Permutation and Combination
119280
If ' \(n\) ' is a positive integer, then \(2.4^{2 n+1}+3^{3 n+1}\) is divisible by
1 2
2 9
3 11
4 27
Explanation:
C : Given that, \(\mathrm{n}\) is a positive integer If \(\mathrm{n}=1\) \(2.4^{2(1)+1}+3^{3(1)+1}\) \(=2.4^3+3^4\) \(=2.64+3^4\) \(=2.64+81=128+81=209\) Hence, 209 is the divisible by 11 .
119277
If \({ }^n p_4=20 \times{ }^n p_2\). Then the value of \(n\) is
1 18
2 13
3 7
4 4
Explanation:
C We have \({ }^n p_4=20 \times{ }^n p_2\) \(\frac{n !}{(n-4) !}=20 \times \frac{n !}{(n-2) !}\) \(\frac{1}{(n-4) !}=20 \times \frac{1}{(n-2)(n-3)(n-4) !}\) \((n-2)(n-3)=20\) \(n^2-3 n-2 n+6=20\) \(n^2-5 n-14=0\) \((n-7)(n+2)=0\) \(n=7\)
AMU-2017
Permutation and Combination
119279
The number of words which can be made out of the letters of the word 'MOBILE' when consonants occupy odd places is
1 20
2 36
3 30
4 720
Explanation:
B Given that, \(\text { Total word in MOBILE }=6 !\) \(\text { Number of consonants }=\mathrm{M}, \mathrm{B}, \mathrm{L}=3\) \(\text { Number of vowel = O,I,E = 3 }\) At odd places 3 consonants take place in 3! ways. and remaining places 3 vowel can take place in 3 ! ways. thus required words \(=3 ! \times 3 !=6 \times 6=36\)
APEAPCET- 23.08.2021
Permutation and Combination
119280
If ' \(n\) ' is a positive integer, then \(2.4^{2 n+1}+3^{3 n+1}\) is divisible by
1 2
2 9
3 11
4 27
Explanation:
C : Given that, \(\mathrm{n}\) is a positive integer If \(\mathrm{n}=1\) \(2.4^{2(1)+1}+3^{3(1)+1}\) \(=2.4^3+3^4\) \(=2.64+3^4\) \(=2.64+81=128+81=209\) Hence, 209 is the divisible by 11 .
