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+9 votes

$m$ identical balls are to be placed in $n$ distinct bags. You are given that $m \geq kn$, where $k$ is a natural number $\geq 1$. In how many ways can the balls be placed in the bags if each bag must contain at least $k$ balls?

  1. \(\left( \begin{array}{c} m - k \\ n - 1 \end{array} \right)\)
  2. \(\left( \begin{array}{c} m - kn + n - 1 \\ n - 1 \end{array} \right)\)
  3. \(\left( \begin{array}{c} m - 1 \\ n - k \end{array} \right)\)
  4. \(\left( \begin{array}{c} m - kn + n + k - 2 \\ n - k \end{array} \right)\)
asked in Combinatory by Veteran (69k points)
retagged by | 1.1k views

2 Answers

+18 votes
Best answer
As there have to be atleast $k$ balls in each bag, so firstly put $k$ balls in each bag i.e $\left(k*n\right)$ balls.

Then now we have total $\left(m-k*n\right)$ balls remaining.

We can use balls $\&$ sticks method now $!$

$n$ bags$= n$ variables, they need to equal to $\left(m-k*n\right)$, no restrictions on how many balls in each bag $!$

$x_{1}+x_{2}+\cdots+x_{n}= \left ( m-k*n \right ),x_1,x_2\cdots x_n>=0.$

So when we solve it

We get

$C(m - k*n + n - 1, n-1 ) = C(m - k*n + n - 1, m- k*n )$
answered by Veteran (49.5k points)
edited by
Can you explain the ball and sticks method? How did you get the final answer. Please give some reference.
x1 + x2 + ... + xn = m- k*n , x1,x2..xn >=0.

How do we solve this???
One way to solve it

there are m balls

each bag must contain atleast k balls

So, after giving each bag k balls , there remain m-kn balls

now in remaining these balls we need to distribute in all bags by distribution method

C(m-kn+n-1, n-1) ways

Hey @Akash Kanase, if the balls were to be distinct then will this method still be applied. I am asking in reference to a question: , which is similar to the one answered above. Can you please explain? 

+9 votes

As there have to be atleast k balls in each bag, so firstly put k balls in each bag i.e k*n balls. Then after, (m - kn) identical balls are left which we have to put it in n distinct bags, so use this general formula: 

C(n + m - kn - 1, n -1).

So, answer would be b.

answered by Loyal (2.5k points)

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