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1.

Statement-1 : The number of ways of distributing 10 identical balls in 4 distinctboxes such that no box is empty is `^9C_3`.Statement-2 : The number of ways of choosing any 3 places from 9 different places is`^9C_3`.Statement-1 is true, Statement-2 is true; Statement-2 is a correctexplanation for Statement-1.Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation forStatement-1.Statement-1 is true, Statement-2 is false.Statement-1 is false, Statement-2 is true.

Answer» 1) total ways = `.^(n+r-1)C_(n-1)`
`= .^(6+4-1)C_(4-1) = .^9C_3`
2)`.^9C_3`
option 3 is correct
Discussion
2.

There are m men and n monkeys (n > m). If a man have any number of monkeys. In how many ways may every monkey have a master?

Answer» The first monkey can select his master by m ways and after that the second monkey can select his master again by m ways, so can the third and so on.
all monkeys can select master `=mxxmxxm`. .. . Upto n factors `=(m)^(n)` ways
Discussion
3.

In how many ways can 4 prizes be distributed among 5 students, if no student gets all the prizes?

Answer» The number of ways in which the 4 prizes can be given away to the 5 students, if a student can get any number of prizes `=5^(4)=625`
Again, the number of ways in which a student gets all the 4 prizes=5, since there are 5 students and any one of them may get all the four prizes.
therefore, the required number of ways in which a student does not get all the prizes=625-5=620.
Discussion
4.

How many four-digit numbers can be formed by using the digits 1, 2, 3,4, 5, 6, 7 if at least one digit is repeated.

Answer» The numbers that can be formed when repetition of digits is allowed are `7^(4)=2401`.
the number that can be formed when all the digits are distincct when repetition is not allowed are `.^(7)P_(4)=840`.
therefore, the numbers that can be formed when atleast one digit is repeated `=7^(4)-.^(7)P_(4)`
`=2401-840=1561`
Discussion
5.

The number of 6 digits numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digit is repeated isA. 60B. 72C. 48D. 36

Answer» Correct Answer - A
Key idea Use divisibility test of 11 and consider different situation according to given condition.
Since, the sum of given digits
0 + 1 + 2 + 5 + 7 + 9 = 24
Let the six - digit number be abcdef and to be divisible by 11, so the difference of sum of odd placed digits and sum of even placed digits should be either 0 or a multiple of 11 means `| (a + c + e) - (b + d + f)|` should be either 0 or a multiple of 11.
Hence possible case is a + c + e = 12 = b + d + f (only)
Now, Case I
set {a, c, e} = {0, 5, 7} and set {b, d, f} = {1, 2, 9}
So, number of 6 - digits number `= (2 xx 2!) xx (3!) = 24`
[`because` a can be selected in ways only either 5 or 7].
Case II
Set {a, c, e} = {1, 2, 9} and set {b, d, f} = {0, 5, 7}
So, number of 6 - digits number `= 3! xx 3! = 36`
So, total number of 6 - digits numbers = 24 + 36 = 60
Discussion