118985
In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?
1 400
2 700
3 1050
4 2100
Explanation:
D Given that, number of students \(=10\) We know that, the number of ways of dividing \((m+n+p)\) things into three groups containing \(\mathrm{m}, \mathrm{n}\) and \(\mathrm{p}\) things respectively \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} !}\) Further if any two groups out of the three have same number of things then number of ways \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} ! \times 2}\) Therefore, number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three \(=\frac{10 !}{4 ! 3 ! 3 ! \times 2}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 3 \times 2 \times 2}=2100\)
VITEEE-2008
Permutation and Combination
118986
Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
1 9
2 12
3 10
4 14
Explanation:
B Here, let there are \(\mathrm{n}\) persons in the room. The total number of hand shakes is some as the number of ways of selecting 2 out of \(n\). \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=66\) \(\Rightarrow \quad \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=66\) \(\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2 !}=66\) \(\Rightarrow \quad \mathrm{n}^2-\mathrm{n}-132=0\) \(\Rightarrow \quad(\mathrm{n}-12)(\mathrm{n}+11)=0\) \(\mathrm{n}=12\)
Manipal UGET-2018]
Permutation and Combination
118987
The number of positive integral solutions of the equation \(x_1 x_2 x_3 x_4 x_5=1050\) is
1 1870
2 1875
3 1865
4 1880
Explanation:
B Given that, \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=1050\) \(\Rightarrow \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=2 \times 3 \times 5^2 \times 7\) Each of 2,3 or 7 can take 5 places and \(5^2\) can be assign in 15 ways. \(\text { Hence, number of positive integral solution } =5^3 \times 15\) \(=1875\)
VITEEE-2007
Permutation and Combination
118989
Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
1 69760
2 30240
3 99784
4 None of these
Explanation:
A : Here, total number of words that can be formed \(=10^5\) Total number of words without repeat in \(={ }^{10} \mathrm{P}_5\) \(\therefore\) Words that have at least one letter is repeated is \(=10^5-{ }^{10} \mathrm{P}_5\) \(=100000-30240\) \(=69760\)
118985
In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?
1 400
2 700
3 1050
4 2100
Explanation:
D Given that, number of students \(=10\) We know that, the number of ways of dividing \((m+n+p)\) things into three groups containing \(\mathrm{m}, \mathrm{n}\) and \(\mathrm{p}\) things respectively \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} !}\) Further if any two groups out of the three have same number of things then number of ways \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} ! \times 2}\) Therefore, number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three \(=\frac{10 !}{4 ! 3 ! 3 ! \times 2}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 3 \times 2 \times 2}=2100\)
VITEEE-2008
Permutation and Combination
118986
Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
1 9
2 12
3 10
4 14
Explanation:
B Here, let there are \(\mathrm{n}\) persons in the room. The total number of hand shakes is some as the number of ways of selecting 2 out of \(n\). \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=66\) \(\Rightarrow \quad \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=66\) \(\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2 !}=66\) \(\Rightarrow \quad \mathrm{n}^2-\mathrm{n}-132=0\) \(\Rightarrow \quad(\mathrm{n}-12)(\mathrm{n}+11)=0\) \(\mathrm{n}=12\)
Manipal UGET-2018]
Permutation and Combination
118987
The number of positive integral solutions of the equation \(x_1 x_2 x_3 x_4 x_5=1050\) is
1 1870
2 1875
3 1865
4 1880
Explanation:
B Given that, \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=1050\) \(\Rightarrow \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=2 \times 3 \times 5^2 \times 7\) Each of 2,3 or 7 can take 5 places and \(5^2\) can be assign in 15 ways. \(\text { Hence, number of positive integral solution } =5^3 \times 15\) \(=1875\)
VITEEE-2007
Permutation and Combination
118989
Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
1 69760
2 30240
3 99784
4 None of these
Explanation:
A : Here, total number of words that can be formed \(=10^5\) Total number of words without repeat in \(={ }^{10} \mathrm{P}_5\) \(\therefore\) Words that have at least one letter is repeated is \(=10^5-{ }^{10} \mathrm{P}_5\) \(=100000-30240\) \(=69760\)
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Permutation and Combination
118985
In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?
1 400
2 700
3 1050
4 2100
Explanation:
D Given that, number of students \(=10\) We know that, the number of ways of dividing \((m+n+p)\) things into three groups containing \(\mathrm{m}, \mathrm{n}\) and \(\mathrm{p}\) things respectively \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} !}\) Further if any two groups out of the three have same number of things then number of ways \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} ! \times 2}\) Therefore, number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three \(=\frac{10 !}{4 ! 3 ! 3 ! \times 2}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 3 \times 2 \times 2}=2100\)
VITEEE-2008
Permutation and Combination
118986
Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
1 9
2 12
3 10
4 14
Explanation:
B Here, let there are \(\mathrm{n}\) persons in the room. The total number of hand shakes is some as the number of ways of selecting 2 out of \(n\). \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=66\) \(\Rightarrow \quad \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=66\) \(\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2 !}=66\) \(\Rightarrow \quad \mathrm{n}^2-\mathrm{n}-132=0\) \(\Rightarrow \quad(\mathrm{n}-12)(\mathrm{n}+11)=0\) \(\mathrm{n}=12\)
Manipal UGET-2018]
Permutation and Combination
118987
The number of positive integral solutions of the equation \(x_1 x_2 x_3 x_4 x_5=1050\) is
1 1870
2 1875
3 1865
4 1880
Explanation:
B Given that, \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=1050\) \(\Rightarrow \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=2 \times 3 \times 5^2 \times 7\) Each of 2,3 or 7 can take 5 places and \(5^2\) can be assign in 15 ways. \(\text { Hence, number of positive integral solution } =5^3 \times 15\) \(=1875\)
VITEEE-2007
Permutation and Combination
118989
Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
1 69760
2 30240
3 99784
4 None of these
Explanation:
A : Here, total number of words that can be formed \(=10^5\) Total number of words without repeat in \(={ }^{10} \mathrm{P}_5\) \(\therefore\) Words that have at least one letter is repeated is \(=10^5-{ }^{10} \mathrm{P}_5\) \(=100000-30240\) \(=69760\)
118985
In how many number of ways can 10 students be divided into three teams, one containing four students and the other three?
1 400
2 700
3 1050
4 2100
Explanation:
D Given that, number of students \(=10\) We know that, the number of ways of dividing \((m+n+p)\) things into three groups containing \(\mathrm{m}, \mathrm{n}\) and \(\mathrm{p}\) things respectively \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} !}\) Further if any two groups out of the three have same number of things then number of ways \(=\frac{(\mathrm{m}+\mathrm{n}+\mathrm{p}) !}{\mathrm{m} ! \mathrm{n} ! \mathrm{p} ! \times 2}\) Therefore, number of ways to divide 10 students into three teams one containing four students and each remaining two teams contain three \(=\frac{10 !}{4 ! 3 ! 3 ! \times 2}=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 3 \times 2 \times 2}=2100\)
VITEEE-2008
Permutation and Combination
118986
Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
1 9
2 12
3 10
4 14
Explanation:
B Here, let there are \(\mathrm{n}\) persons in the room. The total number of hand shakes is some as the number of ways of selecting 2 out of \(n\). \(\therefore \quad{ }^{\mathrm{n}} \mathrm{C}_2=66\) \(\Rightarrow \quad \frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}=66\) \(\Rightarrow \frac{\mathrm{n}(\mathrm{n}-1)}{2 !}=66\) \(\Rightarrow \quad \mathrm{n}^2-\mathrm{n}-132=0\) \(\Rightarrow \quad(\mathrm{n}-12)(\mathrm{n}+11)=0\) \(\mathrm{n}=12\)
Manipal UGET-2018]
Permutation and Combination
118987
The number of positive integral solutions of the equation \(x_1 x_2 x_3 x_4 x_5=1050\) is
1 1870
2 1875
3 1865
4 1880
Explanation:
B Given that, \(\mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=1050\) \(\Rightarrow \mathrm{x}_1 \mathrm{x}_2 \mathrm{x}_3 \mathrm{x}_4 \mathrm{x}_5=2 \times 3 \times 5^2 \times 7\) Each of 2,3 or 7 can take 5 places and \(5^2\) can be assign in 15 ways. \(\text { Hence, number of positive integral solution } =5^3 \times 15\) \(=1875\)
VITEEE-2007
Permutation and Combination
118989
Ten different letters of an alphabet are given, words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is
1 69760
2 30240
3 99784
4 None of these
Explanation:
A : Here, total number of words that can be formed \(=10^5\) Total number of words without repeat in \(={ }^{10} \mathrm{P}_5\) \(\therefore\) Words that have at least one letter is repeated is \(=10^5-{ }^{10} \mathrm{P}_5\) \(=100000-30240\) \(=69760\)