The required number of words is : (2C1 x 4C2 + 2C2 x 2C1) 3! = 96
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
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
Volumes of the same book are not to be separated i, e. all volumes of the same book are to be kept together, Regarding all volumes of the same book as one book, we have only 4 + 1 + 1 + 1 = 7 books.
These seven books can be arranged in 7! ways. The book having 8 volumes can be arranged among themselves in 8! ways, the book having 5 volumes can be arranged among themselves in 5! ways, and the book having 3 volumes can be arranged among themselves in 3! ways.
? Required number = 7! 8! 5! 3!
Possible arrangements are 7!/2! 2! = 1260
[Division by 2 times 2! is because of the repletion of E and R .]
Taping all vowels (IEO) as a single letter (since they come together ) there are six letters with two 'R' s
Hence no. of arrangement = 6!/2! x 3! = 2160
[3 vowels can be arranged in 3! ways among themselves, here multiplied with 3!. ]
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
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
Total number of password using all alphabets -Total
number of password using no symmetric alphabets
= (26 x 25 x 24 ) - (15 x 14 x 13 )
= 12870
Each question can be answered in 2 ways.
? 10 question can be answered = 210 = 1024 ways
Maximum number of such different groups = ABC , ABD, ABE, BCE, BDE, CEA, DEA = 7
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