In how many ways can 9 identical balls be distributed among four

Question

In how many ways can 9 identical balls be distributed among four baskets such that each basket gets at least one ball?

A. 35
B. 56
C. 63
D. 70
E. 126

GMATinsight · Accepted Answer

This question is based on Distribution principle  (CONCEPT VIDEO Attached) . It can be done in two ways

This is like find total Natural number solution of an equation a+b+c+d = 9

Method 1:

Natural Number solution of an equation in r variables =  ^{(n-1)}C_{(r-1)}  =  ^{(9-1)}C_{(4-1)}  =  ^8C_3  = 56

Method 2:

Assign one ball to each basket i.e. 4 balls are gone and we are left with only 5 balls to assign to the baskets
Now every basket needs balls from 0 to 5
a+b+c+d = 5
and a, b, c and d may be any integer from 0 to 5
Now manually find solutions which is long but you get a pattern in that as well leading you to the answer i,e, 56

Answer: option B

CONCEPT VIDEO: 
 https://www.youtube.com/watch?v=I2XaJ2qEV1E

CrackverbalGMAT · Answer

There are 2 important equations for these types of arrangements

If the number of non negative integer solutions for the equation  x_1 + x_2 \space + \space ..+ \space x_n \space = \space n , then the number of ways the distribution can be done is:

(i)  ^{n+r−1}C_{r−1} . In this case, value of any variable can be zero.

(ii)  ^{n−1}C_{r−1} . In this case, minimum value for any variable is 1.

Given that: n = 9 balls, r = 4 boxes and and each box should have at least one ball.

Therefore the total number of ways =  ^{n-1}C_{r-1} = \space ^{9-1}C_{4-1} = \space ^8C_3 = \frac{8 \space * \space 7 \space * \space 6}{3 \space * \space 2 \space * \space 1} = 56

Option B

Arun Kumar

pushpitkc · Answer

If we were to manually list down all the possibilities for the four baskets(and the 4 digit number is the total number of balls in the basket number one, two, three, and four)

6-1-1-1 (4 ways) - 6111,1611,1161,1116
3-2-2-2 (4 ways) - 3222,2322,2232,2223
4-2-2-1 (12 ways) - 4221,4212,4122,2124,2142,2421,2412,2241,2214,1224,1242,1422
4-3-1-1 (12 ways) - 4311,4131,4113,3411,3114,3141,1341,1431,1143,1134,1413,1314
1-2-3-3 (12 ways) - 1233,1323,1332,2133,2331,2313,3321,3231,3123,3132,3312,3213
5-1-1-2 (12 ways) - 5112,5121,5211,2511,2151,2115,1215,1251,1125,1152,1521,1512

Therefore, there are  4*2 + 12*4  or  56 ways(Option B)  in which the 9 balls can be arranged among the 4 baskets.

chetan2u · Answer

hi..

the question does NOT mention about the baskets, if they too are identical..
  1) Identical boxes..  
a) 6,1,1,1 
b) 5,2,1,1
c) 4,3,1,1
d) 4,2,2,1
e) 3,3,2,1
f) 3,2,2,2
so 6 ways

2) non identical boxes..  
a) 6,1,1,1 .............  \frac{4!}{3!} = 4 
b) 5,2,1,1.............  \frac{4!}{2!} = 4*3=12 
c) 4,3,1,1.............  \frac{4!}{2!} = 4*3=12 
d) 4,2,2,1.............  \frac{4!}{2!} = 4*3=12 
e) 3,3,2,1.............  \frac{4!}{2!} = 4*3=12 
f) 3,2,2,2.............  \frac{4!}{3!} = 4 
so 4+12+12+12+12+4=56
In these cases we can take it as an equation a+b+c+d=9, as shown above..

For other type of such possible scenarios
 https://gmatclub.com/forum/topic215915.html

Anuragjn · Answer

As Baskets are 4, solution should be a multiple of 4. Instead of calculating it the long way, we can use Triage in this question.

Only solution multiple of 4 is B. Hence the Answer.

RitiMisra · Answer

Hi  Bunuel ,
Can you please help us with an easy way of solving these kinds of problems ( and some more examples maybe ?)
Thanks in advance.

ShankSouljaBoi · Answer

Distribute 1 ball to each basket.
Now how many balls remain ? 5
How many baskets ? 4

Technique 1
n+r-1 C r-1 n= 5 r= 4 = 8C3

Technique 2

4 baskets can be divided with three separators 
A S1 B S2 C S3 D 
Now 5 balls and 3 separators can be permuted in (5+3)! ways
But separators and balls are identical , hence divide (5+3)! By 3!*5!
Final answer (5+3)! / 3!*5!

B

GMATinsight · Answer

Hi Hoozan Pardon my intrusion but here is the concept video of the partition rule concept. I hope that you find it useful

https://www.youtube.com/watch?v=I2XaJ2qEV1E

Akshaynandurkar · Answer

Hi
there are 4 baskets and each basket requires at least 1 ball. So distribute those 4 balls first in these baskets. Now we are remaining with 5 balls. This can be distributed as follows:
1) 1 1 1 2 ---> 4!/3! = 4 ways
2) 0 0 0 5 ---->4!/3! = 4 ways
3) 0 1 0 4 ----->4!/2! = 12 ways 
see the pattern here if we continue like this add them at last we will get the number that is multiple of 4. Only option B satisfies that condition. remaining can be eliminated. This is how I solve this question in less than a 2 minute. But I took the advantage of options.

Su1206 · Answer

How will the solution change if there have to be atleast 2 balls in each basket? Please help

Bunuel · Answer

If each basket must have  at least 2 balls , then start by putting 2 balls in each basket. That uses up 8 balls, leaving 1 ball. That last ball can go into any of the 4 baskets, so the answer in that case is  4 ways .

Now, note that the original problem (at least 1 ball in each basket) can be solved without convoluted formulas. Give 1 ball to each basket, which leaves 5 balls to distribute. Then just list the cases:

{5 0 0 0}, which gives 4!/3! = 4 permutations 
 {4 1 0 0}, which gives 4!/2! = 12 permutations 
 {3 2 0 0}, which gives 4!/2! = 12 permutations 
 {3 1 1 0}, which gives 4!/2! = 12 permutations 
 {2 2 1 0}, which gives 4!/2! = 12 permutations 
 {2 1 1 1}, which gives 4!/3! = 4 permutations

Adding them up gives  56 .

So the answers are:

With “at least 1 ball”: 56 
 With “at least 2 balls”: 4

Rickooreoisb · Answer

Formula driven

number of ways in which each value has to be atleast one (this question)

n-1Cr-1

Extra : Number of ways in when zero are also allowed

n+r-1 C r-1

In how many ways can 9 identical balls be distributed among four

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