in how many ways can 9 identical balls be distributed among four baske

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

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
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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.

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
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chetan2u

If we remove the last condition i.e. such that each basket gets at least one ball. The possible combination - 9!^4 am I correct?

Could you please explain what "natural number solution" means?

How would the approach change if the numbers changed?

e.g. if 10 identical balls were distributed among 4 boxes and each box gets at least 2 balls, in how many ways could the balls be distributed?

Posted from my mobile device

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.

Q.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

Correct Answere is B

Anuragjn reasoning seems wrong to me, as we are summing different numbers of 4 baskets (not multiplying), so not necessarily divisible by 4

Can someone please explain?
Is this question within scope of gmat?

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.

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

Kudo, if this helped.

Posted from my mobile device
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Asked: In how many ways can 9 identical balls be distributed among four baskets such that each basket gets at least one ball?

B|B|B|B

4 identical balls are placed in 4 baskets. Remaining 5 identical balls are to be placed in 4 baskets
Number of ways = 8!/5!3! = 56

IMO B

GMATBusters could you explain this one

you can read the concept here: https://brilliant.org/wiki/identical-ob ... inct-bins/

We will discuss the same in our session.


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Hi Hoozan

Pardon my intrusion but here is the concept video of the partition rule concept.

I hope that you find it useful



Get TOPICWISE: Concept Videos | Practice Qns 100+ | Official Qns 50+ | 100% Video solution CLICK.
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Thank you

Posted from my mobile device

IMO B

Since each basket should have at least one ball let's fulfil that condition first by giving one ball to each basket.
Now the remaining balls can be distributed in any manner with any basket receiving any number of balls between 0,5 both inclusive .
Therefore we use the formula

n+r-1 C r-1

where n=5 ,r =4

8Cr =56

Two ways we can solve this.
Method 1: All balls are identical so take any 4 balls and put one ball in each basket at first because the question says each basket shall get at least one ball. This is done in 1 way. Remaining 9-4=5 balls. These 5 identical balls now have to be put in 4 baskets in
(n+r-1)C(r-1) ways where n=5 (balls) and r=4 (baskets). So, (5+4-1)C(4-1)=8C3=56 ways(B).

Method 2: Each basket shall have at least one ball. So, (n-1)C(r-1). Here, n=9 (balls) and r=4 (baskets). (9-1)C(4-1)=8C3=56 ways(B).