E-StoreHome » Ask & Discuss » Mathematics. » Algebra « Back to Discussion
Algebra
Comments (125)
Wow!! just look at hte number of views for this problem
Ok people here is my approach to solve this problem:
We are suppose to find the number of ways to distribute 30 identical balls in 3 Identical boxes none empty
Now consider dis dummy problem no. positive integral of solutions of x+y+z = 30
Answer of this dummy prob. is 29C2 = 406.
But wait not so fast 406 will definitely not be the answer for our main question because in the dummy problem x, y, z were distinct.
For us x, y, z are identical.
so what to do???
Out of 406 solutions 1 corresponds to the solution {10,10,10} which has been counted only once.
Now we are remaining with other 405 solutions.
Out of theses 405 solutions there are cases like {5,5,20}, {5,20,5}, {20,5,5,} in general two same solutions which have been counted
3 times. (let number of such unique solutions be i )
Also there solutions like {12,8,10}, {12,10,8}, {8,10,12}, {8,12,10},.... in general all three different which have benn counted
3! = 6times (let number of such unique solutions be j)
so 3i + 6j = 405
Now to find no. of solutions in which any two of x, y, z are same
i.e. to find no. of non-negative solutions of 2g + h =27 .................(no box should be empty so give them each 1 ball does not matter as all balls are identical)
which are clearly 13 ......................... g can vary from 0 to 13 but g = 9 is the set {10,10,10} which has been taken care of.
so i=13 .............no. of sol. in which two are equal
now 3*13 + 6j = 405 hence j = 61
Hence the total number of solutions are
all same + two same + no same (all different ) = 1 + i + j = 1 + 13 + 61 = 75
Hence the answer is 75!!!
yup, but this one is easier one!!
case with (look down) no. of such possible cases
28 balls in one box & rest in others= 1
27 in one = 1
26 in one = 2
25 in one = 2
24 in one = 3
23 in one = 3
22 in one = 4
21 in one = 4
20 in one = 5
19 in one = 5
18 in one = 6
17 in one = 6
16 in one = 7
15 in one = 7
14 in one= 8-1 7 ( one counted in above case)
13 in one = 8-3 5 (similar reason)
12 in one= 9-5 4
11 in one = 9-7 2
10 in one = 10-9 1 stop, no more cases possible
these all add up to 75, this one is easy i think.....
sorry, but here's it gain, i want to know the right answer, so here's my trial
case with (look down) no. of such possible cases
28 balls in one box & rest in others= 1
27 in one = 1
26 in one = 2
25 in one = 2
24 in one = 3
23 in one = 3
22 in one = 4
21 in one = 4
20 in one = 5
19 in one = 5
18 in one = 6
17 in one = 6
16 in one = 7
15 in one = 7
14 in one= 8-1 7 ( one counted in above case)
13 in one = 8-3 5 (similar reason)
12 in one= 9-5 4
11 in one = 9-7 2
10 in one = 10-9 1 stop, no more cases possible
these all add up to 75, this one is easy i think.....
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
Find Posts by Topics
Physics.
TopicsMathematics.
Chemistry.
Biology
Institutes
Parents
Board
Fun Zone
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
2725 Students Currently Online |
Connect with Us  |
 |
56.
here's how.
a minimum of a ball per box. let box B and C have 1 ball each. then box A has 28. keep on decreasing a ball from box A and figure the combinations in boxes B and C. the total is 1+1+2+2+3+3+4+4+5+5+6+6+7+7.