118970
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
1 1200
2 1220
3 1240
4 1260
Explanation:
D Total number of discs \((\mathrm{n})=(4+3+2)\) \(\mathrm{n}=9\) First kind of red colour \(=4\) Second kind of yellow \(=3\) Third kind of Green \(=2\) \(\therefore\) The total number of arrangement \(=\frac{n !}{P_{1} ! P_{2} ! P_{3} !}\) \(=\frac{9 !}{4 ! 3 ! 2 !}\) \(=\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1}\) \(=1260\)
COMEDK-2017
Permutation and Combination
118971
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
1 \(10 ! \times 2\)
2 10 !
3 \(9 ! \times 2\)
4 \(11 ! \times 2\) !
Explanation:
C Consider the president and secretary as one member. So, the total members are 10 (11-1). These 10 member can be seat around the circular table \(=(10-1)\) ! \(=9\) ! Two member can sit in two ways \(=2\) So, the total number of ways \(=9 ! \times 2\)
COMEDK-2018
Permutation and Combination
118972
Seven different letters are given, then the number of ways in which words of 5 letters can be formed such that at least one of the letters is repeated is
1 \({ }^7 \mathrm{P}_5\)
2 14287
3 \(5^7\)
4 \(7^5\)
Explanation:
B : Given that, number of letter \(=7\) Formed words \(=5\) So, the required number of ways- \(=7^5{ }^7 \mathrm{P}_5\) \(=16807-2520\) \(=14287\)
COMEDK-2020
Permutation and Combination
119050
The number of diagonals in a polygon is 20 . The number of sides of the polygon is
1 5
2 6
3 8
4 10
Explanation:
C We know that, the formula to find the number \(\text { of diagonals }=\frac{n(n-3)}{2}\) \(20=\frac{n(n-3)}{2}\) \(40=n^2-3 n\) \(n^2-3 n-40=0\) \(n^2-8 n+5 n-40=0\) \((n-8)(n+5)=0\) \(n=8\)
WB JEE-2011
Permutation and Combination
118961
The number of ways to arrange the letters of word CHEESE are:
1 120
2 240
3 720
4 6
Explanation:
A We have to find the total number of arrangements in words CHEESE Total words \(=6\) Repeated word \(=3\) So, the required number of ways \(=\frac{6 !}{3 !}\) \(=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=6 \times 5 \times 4\) \(=120\)
118970
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
1 1200
2 1220
3 1240
4 1260
Explanation:
D Total number of discs \((\mathrm{n})=(4+3+2)\) \(\mathrm{n}=9\) First kind of red colour \(=4\) Second kind of yellow \(=3\) Third kind of Green \(=2\) \(\therefore\) The total number of arrangement \(=\frac{n !}{P_{1} ! P_{2} ! P_{3} !}\) \(=\frac{9 !}{4 ! 3 ! 2 !}\) \(=\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1}\) \(=1260\)
COMEDK-2017
Permutation and Combination
118971
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
1 \(10 ! \times 2\)
2 10 !
3 \(9 ! \times 2\)
4 \(11 ! \times 2\) !
Explanation:
C Consider the president and secretary as one member. So, the total members are 10 (11-1). These 10 member can be seat around the circular table \(=(10-1)\) ! \(=9\) ! Two member can sit in two ways \(=2\) So, the total number of ways \(=9 ! \times 2\)
COMEDK-2018
Permutation and Combination
118972
Seven different letters are given, then the number of ways in which words of 5 letters can be formed such that at least one of the letters is repeated is
1 \({ }^7 \mathrm{P}_5\)
2 14287
3 \(5^7\)
4 \(7^5\)
Explanation:
B : Given that, number of letter \(=7\) Formed words \(=5\) So, the required number of ways- \(=7^5{ }^7 \mathrm{P}_5\) \(=16807-2520\) \(=14287\)
COMEDK-2020
Permutation and Combination
119050
The number of diagonals in a polygon is 20 . The number of sides of the polygon is
1 5
2 6
3 8
4 10
Explanation:
C We know that, the formula to find the number \(\text { of diagonals }=\frac{n(n-3)}{2}\) \(20=\frac{n(n-3)}{2}\) \(40=n^2-3 n\) \(n^2-3 n-40=0\) \(n^2-8 n+5 n-40=0\) \((n-8)(n+5)=0\) \(n=8\)
WB JEE-2011
Permutation and Combination
118961
The number of ways to arrange the letters of word CHEESE are:
1 120
2 240
3 720
4 6
Explanation:
A We have to find the total number of arrangements in words CHEESE Total words \(=6\) Repeated word \(=3\) So, the required number of ways \(=\frac{6 !}{3 !}\) \(=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=6 \times 5 \times 4\) \(=120\)
118970
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
1 1200
2 1220
3 1240
4 1260
Explanation:
D Total number of discs \((\mathrm{n})=(4+3+2)\) \(\mathrm{n}=9\) First kind of red colour \(=4\) Second kind of yellow \(=3\) Third kind of Green \(=2\) \(\therefore\) The total number of arrangement \(=\frac{n !}{P_{1} ! P_{2} ! P_{3} !}\) \(=\frac{9 !}{4 ! 3 ! 2 !}\) \(=\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1}\) \(=1260\)
COMEDK-2017
Permutation and Combination
118971
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
1 \(10 ! \times 2\)
2 10 !
3 \(9 ! \times 2\)
4 \(11 ! \times 2\) !
Explanation:
C Consider the president and secretary as one member. So, the total members are 10 (11-1). These 10 member can be seat around the circular table \(=(10-1)\) ! \(=9\) ! Two member can sit in two ways \(=2\) So, the total number of ways \(=9 ! \times 2\)
COMEDK-2018
Permutation and Combination
118972
Seven different letters are given, then the number of ways in which words of 5 letters can be formed such that at least one of the letters is repeated is
1 \({ }^7 \mathrm{P}_5\)
2 14287
3 \(5^7\)
4 \(7^5\)
Explanation:
B : Given that, number of letter \(=7\) Formed words \(=5\) So, the required number of ways- \(=7^5{ }^7 \mathrm{P}_5\) \(=16807-2520\) \(=14287\)
COMEDK-2020
Permutation and Combination
119050
The number of diagonals in a polygon is 20 . The number of sides of the polygon is
1 5
2 6
3 8
4 10
Explanation:
C We know that, the formula to find the number \(\text { of diagonals }=\frac{n(n-3)}{2}\) \(20=\frac{n(n-3)}{2}\) \(40=n^2-3 n\) \(n^2-3 n-40=0\) \(n^2-8 n+5 n-40=0\) \((n-8)(n+5)=0\) \(n=8\)
WB JEE-2011
Permutation and Combination
118961
The number of ways to arrange the letters of word CHEESE are:
1 120
2 240
3 720
4 6
Explanation:
A We have to find the total number of arrangements in words CHEESE Total words \(=6\) Repeated word \(=3\) So, the required number of ways \(=\frac{6 !}{3 !}\) \(=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=6 \times 5 \times 4\) \(=120\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
118970
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
1 1200
2 1220
3 1240
4 1260
Explanation:
D Total number of discs \((\mathrm{n})=(4+3+2)\) \(\mathrm{n}=9\) First kind of red colour \(=4\) Second kind of yellow \(=3\) Third kind of Green \(=2\) \(\therefore\) The total number of arrangement \(=\frac{n !}{P_{1} ! P_{2} ! P_{3} !}\) \(=\frac{9 !}{4 ! 3 ! 2 !}\) \(=\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1}\) \(=1260\)
COMEDK-2017
Permutation and Combination
118971
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
1 \(10 ! \times 2\)
2 10 !
3 \(9 ! \times 2\)
4 \(11 ! \times 2\) !
Explanation:
C Consider the president and secretary as one member. So, the total members are 10 (11-1). These 10 member can be seat around the circular table \(=(10-1)\) ! \(=9\) ! Two member can sit in two ways \(=2\) So, the total number of ways \(=9 ! \times 2\)
COMEDK-2018
Permutation and Combination
118972
Seven different letters are given, then the number of ways in which words of 5 letters can be formed such that at least one of the letters is repeated is
1 \({ }^7 \mathrm{P}_5\)
2 14287
3 \(5^7\)
4 \(7^5\)
Explanation:
B : Given that, number of letter \(=7\) Formed words \(=5\) So, the required number of ways- \(=7^5{ }^7 \mathrm{P}_5\) \(=16807-2520\) \(=14287\)
COMEDK-2020
Permutation and Combination
119050
The number of diagonals in a polygon is 20 . The number of sides of the polygon is
1 5
2 6
3 8
4 10
Explanation:
C We know that, the formula to find the number \(\text { of diagonals }=\frac{n(n-3)}{2}\) \(20=\frac{n(n-3)}{2}\) \(40=n^2-3 n\) \(n^2-3 n-40=0\) \(n^2-8 n+5 n-40=0\) \((n-8)(n+5)=0\) \(n=8\)
WB JEE-2011
Permutation and Combination
118961
The number of ways to arrange the letters of word CHEESE are:
1 120
2 240
3 720
4 6
Explanation:
A We have to find the total number of arrangements in words CHEESE Total words \(=6\) Repeated word \(=3\) So, the required number of ways \(=\frac{6 !}{3 !}\) \(=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=6 \times 5 \times 4\) \(=120\)
118970
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row, if the discs of the same colour are indistinguishable?
1 1200
2 1220
3 1240
4 1260
Explanation:
D Total number of discs \((\mathrm{n})=(4+3+2)\) \(\mathrm{n}=9\) First kind of red colour \(=4\) Second kind of yellow \(=3\) Third kind of Green \(=2\) \(\therefore\) The total number of arrangement \(=\frac{n !}{P_{1} ! P_{2} ! P_{3} !}\) \(=\frac{9 !}{4 ! 3 ! 2 !}\) \(=\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1}\) \(=1260\)
COMEDK-2017
Permutation and Combination
118971
If eleven members of a committee sit at a round table so that the President and Secretary always sit together, then the number of arrangements is
1 \(10 ! \times 2\)
2 10 !
3 \(9 ! \times 2\)
4 \(11 ! \times 2\) !
Explanation:
C Consider the president and secretary as one member. So, the total members are 10 (11-1). These 10 member can be seat around the circular table \(=(10-1)\) ! \(=9\) ! Two member can sit in two ways \(=2\) So, the total number of ways \(=9 ! \times 2\)
COMEDK-2018
Permutation and Combination
118972
Seven different letters are given, then the number of ways in which words of 5 letters can be formed such that at least one of the letters is repeated is
1 \({ }^7 \mathrm{P}_5\)
2 14287
3 \(5^7\)
4 \(7^5\)
Explanation:
B : Given that, number of letter \(=7\) Formed words \(=5\) So, the required number of ways- \(=7^5{ }^7 \mathrm{P}_5\) \(=16807-2520\) \(=14287\)
COMEDK-2020
Permutation and Combination
119050
The number of diagonals in a polygon is 20 . The number of sides of the polygon is
1 5
2 6
3 8
4 10
Explanation:
C We know that, the formula to find the number \(\text { of diagonals }=\frac{n(n-3)}{2}\) \(20=\frac{n(n-3)}{2}\) \(40=n^2-3 n\) \(n^2-3 n-40=0\) \(n^2-8 n+5 n-40=0\) \((n-8)(n+5)=0\) \(n=8\)
WB JEE-2011
Permutation and Combination
118961
The number of ways to arrange the letters of word CHEESE are:
1 120
2 240
3 720
4 6
Explanation:
A We have to find the total number of arrangements in words CHEESE Total words \(=6\) Repeated word \(=3\) So, the required number of ways \(=\frac{6 !}{3 !}\) \(=\frac{6 \times 5 \times 4 \times 3 !}{3 !}=6 \times 5 \times 4\) \(=120\)