Question

$6$ white and $6$ black balls are distributed among $10$ urns. All the balls are identical except for the color and each urn can hold any number of balls. In how many ways can the distribution be done if at least one ball should be there in each urn?

• A. $12560$
• B. $42150$
• C. $26250$
• D. none of these

Answer 1

The correct option is C $26250$

Case: 1

Select $6$ Urns and place $6$ white balls [or interchange black and white balls]

In rest $4$ place $4$ black balls

Rest $2$ balls can be placed in ${}^{10}{C}_2$ Ways

Number of ways $=\left[{}^{10}{C}_6{\times }^4{C}_4{\times }^{10}{C}_2\right]\times 2$

Case: 2

$6$ White balls in $6$ Urns and $4$ balls in $4$ Urns.

Rest two balls can be placed in same urn. Also revise white and black balls case is also there.

$=\left[{}^{10}{C}_6{\times }^4{C}_4{\times }^{10}{C}_1\right]\times 2$

Case: 3

$5$ White and $5$ black are places in urns $1$ White and $1$ black can be placed in ${}^5{C}_1$ Ways each

${=}^{10}{C}_5{\times }^5{C}_5{\times }^5{C}_1{\times }^5{C}_1$

Therefore total ways $=26250$