Permutation and Combination
- A library has two books each having three copies and three other books each having two copies. In how many ways can all these books be arranged in a shelf so that copies of the same book are not separated ?
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Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.
Correct Option: A
Regarding all copies of the same book as one book, we have only 5 books. These 5 books can be arranged in 5! ways. But all copies of the same book being identical can be arranged in only one way.
∴ Required number = 5! x 1! x 1! x 1! x 1! = 120
- 4 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated ?
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Assume the 2 girl students to be together i,e (one). Now there are 5 students.
Correct Option: D
Assume the 2 girl students to be together i,e (one). Now there are 5 students.
Possible ways of arranging them are 5! = 120
Now they (two girl) can arrange themselves in 2! ways.
Hence, total ways = 120 x 2! = 240
- The total number of words, which can be formed out of the letters a, b, c, d, e, f taken 3 together, such that each word contains at least one vowel is ?
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The required number of words is : (2C1 x 4C2 + 2C2 x 2C1) 3!
Correct Option: C
The required number of words is : (2C1 x 4C2 + 2C2 x 2C1) 3! = 96
- A man has 5 friends and his wife has 4 friends. They want to invite either of their friends, one or more to a party. In how many ways can they do so ?
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Number of ways of selecting one or more friends from 5 friends
= 5C1 + 5C2 + 5C3 + 5C4 + 5C5Correct Option: D
Number of ways of selecting one or more friends from 5 friends
= 5C1 + 5C2 + 5C3 + 5C4 + 5C5
= 5 + 10 + 10 + 5 + 1
= 31 ways
Number of ways of selecting one or more friends from 4 friends = 4C1 + 4C2 + 4C3 + 4C4
= 4 + 6 + 4 + 1
= 15 ways
∴ Total number of ways = 31 + 15 = 46 ways
- If there are 10 positive real numbers n1 < n2, n3 .... < n10 How many triplets of these numbers (n1, n2, n3) (n2, n3, n4), ... can be generated such that in each triplet the first number is always less than the second number and the second number is always less the third number ?
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Three numbers can be selected and arranged out of 10 numbers in 10P3 ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.Correct Option: C
Three numbers can be selected and arranged out of 10 numbers in 10P3 ways 10!/7! = 10 x 9 x 8
Now, this arrangement is restricted to a given condition that first number is always less than the second number and second number is always than the third number. Thus, three numbers can be arranged among themselves in 3! ways.
Hence, required number of arrangement = (10 x 9 x 8)/(3 x 2)
= 120 ways
