119257
The number of ways in which 5 children can be arranged in a row such that two of them Rama and Krishna are always together is
1 48
2 24
3 120
4 72
Explanation:
A Here consider the arrangement by taking two children as one and hence 4 children can now be arranged in \(4 !=24\) ways. Also, two children taken together can be arranged in 2 ways. \(\therefore\) Total no. of ways \(=24 \times 2=48\)
SRM JEEE-2013
Permutation and Combination
119299
How many lines can be drawn from 6 points on a circle?
1 12
2 15
3 30
4 60
Explanation:
B If each line contains two of those points then there are " 6 choose 2 "' possible line \({ }^6 \mathrm{C}_2=\frac{6 !}{2 !(6-2) !}=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}=15\)
Rajasthan PET-2010
Permutation and Combination
119307
\(25^{190}-19^{190}-8^{190}+2^{190}\) is divisible by
1 34 but not by 14
2 both 14 and 34
3 neither 14 not 34
4 14 but not 34
Explanation:
A Given, \(25^{190}-19^{190}-8^{190}+2^{190}\) \(\left(25^{190}-19^{190}\right)-\left(8^{190}-2^{190}\right)\) is divisible by 6 \(\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right)\) is divisible by 17 \(25^{190}-8^{190}\) is not divisible by 7 But \(19^{190}-2^{190}\) is divisible by 7 Hence, \(25^{90}-19^{190}-8^{190}+2^{190}\) is divisible by 34 but not 14
JEE Main-08.04.2023
Permutation and Combination
119264
The number of all three elements subsets of the set \(\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \ldots \mathbf{a}_{\mathbf{n}}\right\}\) which contain \(\mathbf{a}_3\) is
1 \({ }^n C_3\)
2 \({ }^{\mathrm{n}-1} \mathrm{C}_3\)
3 \({ }^{\mathrm{n}-1} \mathrm{C}_2\)
4 None of these
Explanation:
C The number of three elements subsets containing \(a_3\) is equal to the number of ways of selecting 2 elements out of \((n-1)\) elements. So, the required number of subsets is \({ }^{\mathrm{n}-1} \mathrm{C}_2\).
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Permutation and Combination
119257
The number of ways in which 5 children can be arranged in a row such that two of them Rama and Krishna are always together is
1 48
2 24
3 120
4 72
Explanation:
A Here consider the arrangement by taking two children as one and hence 4 children can now be arranged in \(4 !=24\) ways. Also, two children taken together can be arranged in 2 ways. \(\therefore\) Total no. of ways \(=24 \times 2=48\)
SRM JEEE-2013
Permutation and Combination
119299
How many lines can be drawn from 6 points on a circle?
1 12
2 15
3 30
4 60
Explanation:
B If each line contains two of those points then there are " 6 choose 2 "' possible line \({ }^6 \mathrm{C}_2=\frac{6 !}{2 !(6-2) !}=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}=15\)
Rajasthan PET-2010
Permutation and Combination
119307
\(25^{190}-19^{190}-8^{190}+2^{190}\) is divisible by
1 34 but not by 14
2 both 14 and 34
3 neither 14 not 34
4 14 but not 34
Explanation:
A Given, \(25^{190}-19^{190}-8^{190}+2^{190}\) \(\left(25^{190}-19^{190}\right)-\left(8^{190}-2^{190}\right)\) is divisible by 6 \(\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right)\) is divisible by 17 \(25^{190}-8^{190}\) is not divisible by 7 But \(19^{190}-2^{190}\) is divisible by 7 Hence, \(25^{90}-19^{190}-8^{190}+2^{190}\) is divisible by 34 but not 14
JEE Main-08.04.2023
Permutation and Combination
119264
The number of all three elements subsets of the set \(\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \ldots \mathbf{a}_{\mathbf{n}}\right\}\) which contain \(\mathbf{a}_3\) is
1 \({ }^n C_3\)
2 \({ }^{\mathrm{n}-1} \mathrm{C}_3\)
3 \({ }^{\mathrm{n}-1} \mathrm{C}_2\)
4 None of these
Explanation:
C The number of three elements subsets containing \(a_3\) is equal to the number of ways of selecting 2 elements out of \((n-1)\) elements. So, the required number of subsets is \({ }^{\mathrm{n}-1} \mathrm{C}_2\).
119257
The number of ways in which 5 children can be arranged in a row such that two of them Rama and Krishna are always together is
1 48
2 24
3 120
4 72
Explanation:
A Here consider the arrangement by taking two children as one and hence 4 children can now be arranged in \(4 !=24\) ways. Also, two children taken together can be arranged in 2 ways. \(\therefore\) Total no. of ways \(=24 \times 2=48\)
SRM JEEE-2013
Permutation and Combination
119299
How many lines can be drawn from 6 points on a circle?
1 12
2 15
3 30
4 60
Explanation:
B If each line contains two of those points then there are " 6 choose 2 "' possible line \({ }^6 \mathrm{C}_2=\frac{6 !}{2 !(6-2) !}=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}=15\)
Rajasthan PET-2010
Permutation and Combination
119307
\(25^{190}-19^{190}-8^{190}+2^{190}\) is divisible by
1 34 but not by 14
2 both 14 and 34
3 neither 14 not 34
4 14 but not 34
Explanation:
A Given, \(25^{190}-19^{190}-8^{190}+2^{190}\) \(\left(25^{190}-19^{190}\right)-\left(8^{190}-2^{190}\right)\) is divisible by 6 \(\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right)\) is divisible by 17 \(25^{190}-8^{190}\) is not divisible by 7 But \(19^{190}-2^{190}\) is divisible by 7 Hence, \(25^{90}-19^{190}-8^{190}+2^{190}\) is divisible by 34 but not 14
JEE Main-08.04.2023
Permutation and Combination
119264
The number of all three elements subsets of the set \(\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \ldots \mathbf{a}_{\mathbf{n}}\right\}\) which contain \(\mathbf{a}_3\) is
1 \({ }^n C_3\)
2 \({ }^{\mathrm{n}-1} \mathrm{C}_3\)
3 \({ }^{\mathrm{n}-1} \mathrm{C}_2\)
4 None of these
Explanation:
C The number of three elements subsets containing \(a_3\) is equal to the number of ways of selecting 2 elements out of \((n-1)\) elements. So, the required number of subsets is \({ }^{\mathrm{n}-1} \mathrm{C}_2\).
119257
The number of ways in which 5 children can be arranged in a row such that two of them Rama and Krishna are always together is
1 48
2 24
3 120
4 72
Explanation:
A Here consider the arrangement by taking two children as one and hence 4 children can now be arranged in \(4 !=24\) ways. Also, two children taken together can be arranged in 2 ways. \(\therefore\) Total no. of ways \(=24 \times 2=48\)
SRM JEEE-2013
Permutation and Combination
119299
How many lines can be drawn from 6 points on a circle?
1 12
2 15
3 30
4 60
Explanation:
B If each line contains two of those points then there are " 6 choose 2 "' possible line \({ }^6 \mathrm{C}_2=\frac{6 !}{2 !(6-2) !}=\frac{6 \times 5 \times 4 !}{2 \times 1 \times 4 !}=15\)
Rajasthan PET-2010
Permutation and Combination
119307
\(25^{190}-19^{190}-8^{190}+2^{190}\) is divisible by
1 34 but not by 14
2 both 14 and 34
3 neither 14 not 34
4 14 but not 34
Explanation:
A Given, \(25^{190}-19^{190}-8^{190}+2^{190}\) \(\left(25^{190}-19^{190}\right)-\left(8^{190}-2^{190}\right)\) is divisible by 6 \(\left(25^{190}-8^{190}\right)-\left(19^{190}-2^{190}\right)\) is divisible by 17 \(25^{190}-8^{190}\) is not divisible by 7 But \(19^{190}-2^{190}\) is divisible by 7 Hence, \(25^{90}-19^{190}-8^{190}+2^{190}\) is divisible by 34 but not 14
JEE Main-08.04.2023
Permutation and Combination
119264
The number of all three elements subsets of the set \(\left\{\mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \ldots \mathbf{a}_{\mathbf{n}}\right\}\) which contain \(\mathbf{a}_3\) is
1 \({ }^n C_3\)
2 \({ }^{\mathrm{n}-1} \mathrm{C}_3\)
3 \({ }^{\mathrm{n}-1} \mathrm{C}_2\)
4 None of these
Explanation:
C The number of three elements subsets containing \(a_3\) is equal to the number of ways of selecting 2 elements out of \((n-1)\) elements. So, the required number of subsets is \({ }^{\mathrm{n}-1} \mathrm{C}_2\).