The total number of ways of distributing 12 identical balls into 3 DISTINCT boxes would be (12+3-1)c(3-1) = 91
Now, as the boxes are identical, what would be the difference? If I have a distribution of 1,2,9 balls in the boxes, then this specific distribution can be done 3!=6 ways when the boxes are distinct, but in only ONE way, if they are identical.Similarly, if we have a distribution of the type 2,2,8 balls, then this can be done in 3!/2!=3 ways in distinct boxes, but again in only ONE way for identical boxes.If, we have the distribution 4,4,4 balls, then that can be done in only ONE way in both the cases.
So, first I'll distribute the 91 cases into three types:
All distinct number of ballsTwo identical number of ballsAll identical number of ballsWhat we are going to do is, we'll begin with type 2. In this scenario, we have let's say a,a,b balls in the boxes.We have 2a+b=12, whose number of solutions is 6+1=7But out of those 7 cases, 1 is the distribution 4,4,4 which does not fall under this scenario.
So, we have 6 cases under type 3, each of which has been counted thrice, and 1 case of type 3, which has been counted once. And all other are of type 1 which has counted 6 times
Hence distinct possibilities are = 6/3 + 1 + (91-7)/6 = 17This would determine all possible outcomes.
Now, it's easy to find the favourable outcomes3+a+b=12Number of solutions for (a,b) will be (0,9); (1,8) ... (4,5): which is a total of 5 cases
So, probability = 5/17
@gaurav_sharma said in Quant Boosters - Gaurav Sharma - Set 5:
Q26) John and David work on alternate days with John starting on the first day. John does 12.5% of the total work on the last day and finishes the work. If John alone can do the work in 6 days then which of following can be the number of days in which David alone can do the whole work?(a) 24 days(b) 12 days(c) 8 days(d) 10 days
Need solutioninlinemathsinline mathsinlinemaths
Sol: Remember that: If square of a number ends in same unit digit as that of number, then that unit digit can be 0, 1, 5 or 6 only. So possible values for G are 0, 1, 5 or 6. Now by checking you can easily negate that G can't be 0 as GOG will not remain a three digit number. It can't be 1 either as in this case the three digit number will be 1O1 and it'll be perfect square only when O is 2 i.e. T is 1 which is not possible as T and G are "DISTINCT" single digit positive integers. Similarly you can check for 5 also that no number of the form 5O5 is a perfect square.
Only possible case is 26² = 676 i.e. O = 7.