119230
Five points are marked on a circle. The number of distinct polygons of three or more sides can be drawn using some (or all) of the five points as vertices is
1 10
2 12
3 14
4 16
5 18
Explanation:
D Five points with three or more members can be made into exactly one such polygon \(\therefore\) Total subset of five member set \(2^5=32\) 1 have 0 members \(={ }^5 \mathrm{C}_0\) 5 have 1 member \(={ }^5 \mathrm{C}_1\) 10 have 2 member \(={ }^5 \mathrm{C}_2\) Total number of ways \(=32-1-5-10=16\)
Kerala CEE-2020
Permutation and Combination
119231
A box contains 5 red 3 Yellow 4 Green balls if 3 balls are drawn, the number of ways of drawing them so that all of same colour is
1 16
2 15
3 \({ }^5 \mathrm{C}_3\)
4 None of these
Explanation:
B Given that Number of Red balls \(=5\) Number of Yellow balls \(=3\) Number of Green balls \(=4\) All may be red, yellow or green. The required, number of combination is \(={ }^5 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3=10+1+4=15\)
MHT CET-2021
Permutation and Combination
119207
The number of ways in which 3 boys and 2 girls can sit on a bench so that no two boys are adjacent is
1 6
2 10
3 12
4 32
Explanation:
C Given, 3 boys and 2 girls \(\therefore \quad\) No. of ways of girls \(=2\) ! Required number of ways \(=2 ! \times{ }^3 \mathrm{C}_3 \times 3\) ! \(=12\)
AP EAPCET-25.08.2021
Permutation and Combination
119188
How many numbers greater than \(10,00,000\) be formed from 2, 3, 0, 3, 4, 2, 3, ?
1 420
2 360
3 400
4 300
Explanation:
B We know that 10 lakhs has 7 digits in which 2 are similar of first kind and 3 are similar of second kind \(=\frac{7 !}{2 ! 3 !}\) Number of arrangement of the remaining 6 digits \(=\) \(\frac{6 !}{2 ! 3 !}\) \(\text { Total number } =\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\) \(=\frac{7 \times 6 \times 5 \times 4}{2 !}-\frac{6 \times 5 \times 4}{2 !}=420-60=360 .\)
119230
Five points are marked on a circle. The number of distinct polygons of three or more sides can be drawn using some (or all) of the five points as vertices is
1 10
2 12
3 14
4 16
5 18
Explanation:
D Five points with three or more members can be made into exactly one such polygon \(\therefore\) Total subset of five member set \(2^5=32\) 1 have 0 members \(={ }^5 \mathrm{C}_0\) 5 have 1 member \(={ }^5 \mathrm{C}_1\) 10 have 2 member \(={ }^5 \mathrm{C}_2\) Total number of ways \(=32-1-5-10=16\)
Kerala CEE-2020
Permutation and Combination
119231
A box contains 5 red 3 Yellow 4 Green balls if 3 balls are drawn, the number of ways of drawing them so that all of same colour is
1 16
2 15
3 \({ }^5 \mathrm{C}_3\)
4 None of these
Explanation:
B Given that Number of Red balls \(=5\) Number of Yellow balls \(=3\) Number of Green balls \(=4\) All may be red, yellow or green. The required, number of combination is \(={ }^5 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3=10+1+4=15\)
MHT CET-2021
Permutation and Combination
119207
The number of ways in which 3 boys and 2 girls can sit on a bench so that no two boys are adjacent is
1 6
2 10
3 12
4 32
Explanation:
C Given, 3 boys and 2 girls \(\therefore \quad\) No. of ways of girls \(=2\) ! Required number of ways \(=2 ! \times{ }^3 \mathrm{C}_3 \times 3\) ! \(=12\)
AP EAPCET-25.08.2021
Permutation and Combination
119188
How many numbers greater than \(10,00,000\) be formed from 2, 3, 0, 3, 4, 2, 3, ?
1 420
2 360
3 400
4 300
Explanation:
B We know that 10 lakhs has 7 digits in which 2 are similar of first kind and 3 are similar of second kind \(=\frac{7 !}{2 ! 3 !}\) Number of arrangement of the remaining 6 digits \(=\) \(\frac{6 !}{2 ! 3 !}\) \(\text { Total number } =\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\) \(=\frac{7 \times 6 \times 5 \times 4}{2 !}-\frac{6 \times 5 \times 4}{2 !}=420-60=360 .\)
119230
Five points are marked on a circle. The number of distinct polygons of three or more sides can be drawn using some (or all) of the five points as vertices is
1 10
2 12
3 14
4 16
5 18
Explanation:
D Five points with three or more members can be made into exactly one such polygon \(\therefore\) Total subset of five member set \(2^5=32\) 1 have 0 members \(={ }^5 \mathrm{C}_0\) 5 have 1 member \(={ }^5 \mathrm{C}_1\) 10 have 2 member \(={ }^5 \mathrm{C}_2\) Total number of ways \(=32-1-5-10=16\)
Kerala CEE-2020
Permutation and Combination
119231
A box contains 5 red 3 Yellow 4 Green balls if 3 balls are drawn, the number of ways of drawing them so that all of same colour is
1 16
2 15
3 \({ }^5 \mathrm{C}_3\)
4 None of these
Explanation:
B Given that Number of Red balls \(=5\) Number of Yellow balls \(=3\) Number of Green balls \(=4\) All may be red, yellow or green. The required, number of combination is \(={ }^5 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3=10+1+4=15\)
MHT CET-2021
Permutation and Combination
119207
The number of ways in which 3 boys and 2 girls can sit on a bench so that no two boys are adjacent is
1 6
2 10
3 12
4 32
Explanation:
C Given, 3 boys and 2 girls \(\therefore \quad\) No. of ways of girls \(=2\) ! Required number of ways \(=2 ! \times{ }^3 \mathrm{C}_3 \times 3\) ! \(=12\)
AP EAPCET-25.08.2021
Permutation and Combination
119188
How many numbers greater than \(10,00,000\) be formed from 2, 3, 0, 3, 4, 2, 3, ?
1 420
2 360
3 400
4 300
Explanation:
B We know that 10 lakhs has 7 digits in which 2 are similar of first kind and 3 are similar of second kind \(=\frac{7 !}{2 ! 3 !}\) Number of arrangement of the remaining 6 digits \(=\) \(\frac{6 !}{2 ! 3 !}\) \(\text { Total number } =\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\) \(=\frac{7 \times 6 \times 5 \times 4}{2 !}-\frac{6 \times 5 \times 4}{2 !}=420-60=360 .\)
119230
Five points are marked on a circle. The number of distinct polygons of three or more sides can be drawn using some (or all) of the five points as vertices is
1 10
2 12
3 14
4 16
5 18
Explanation:
D Five points with three or more members can be made into exactly one such polygon \(\therefore\) Total subset of five member set \(2^5=32\) 1 have 0 members \(={ }^5 \mathrm{C}_0\) 5 have 1 member \(={ }^5 \mathrm{C}_1\) 10 have 2 member \(={ }^5 \mathrm{C}_2\) Total number of ways \(=32-1-5-10=16\)
Kerala CEE-2020
Permutation and Combination
119231
A box contains 5 red 3 Yellow 4 Green balls if 3 balls are drawn, the number of ways of drawing them so that all of same colour is
1 16
2 15
3 \({ }^5 \mathrm{C}_3\)
4 None of these
Explanation:
B Given that Number of Red balls \(=5\) Number of Yellow balls \(=3\) Number of Green balls \(=4\) All may be red, yellow or green. The required, number of combination is \(={ }^5 \mathrm{C}_3+{ }^3 \mathrm{C}_3+{ }^4 \mathrm{C}_3=10+1+4=15\)
MHT CET-2021
Permutation and Combination
119207
The number of ways in which 3 boys and 2 girls can sit on a bench so that no two boys are adjacent is
1 6
2 10
3 12
4 32
Explanation:
C Given, 3 boys and 2 girls \(\therefore \quad\) No. of ways of girls \(=2\) ! Required number of ways \(=2 ! \times{ }^3 \mathrm{C}_3 \times 3\) ! \(=12\)
AP EAPCET-25.08.2021
Permutation and Combination
119188
How many numbers greater than \(10,00,000\) be formed from 2, 3, 0, 3, 4, 2, 3, ?
1 420
2 360
3 400
4 300
Explanation:
B We know that 10 lakhs has 7 digits in which 2 are similar of first kind and 3 are similar of second kind \(=\frac{7 !}{2 ! 3 !}\) Number of arrangement of the remaining 6 digits \(=\) \(\frac{6 !}{2 ! 3 !}\) \(\text { Total number } =\frac{7 !}{2 ! 3 !}-\frac{6 !}{2 ! 3 !}\) \(=\frac{7 \times 6 \times 5 \times 4}{2 !}-\frac{6 \times 5 \times 4}{2 !}=420-60=360 .\)
