Question

There are two urns. Urn $A$ has $3$ distinct red balls and urn $B$ has $9$ distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is
(A) $36$
(B) $66$
(C) $108$
(D) $3$

Answer 1

Hint: From the given question, it is clear that two different coloured balls are there. By using the combination formula for the two differently coloured balls, we can calculate the combination of two differently coloured balls. The number of ways that the event exists can be calculated by making a product of the combination of the two differently coloured balls and number of balls.

Formula used: The formula used in the theory of combination is,
$n{c_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Where,
$n$ is the number of objects
$c$ is the combination
$r$ is the number of objects to select from set

Complete step-by-step answer:
Total number of balls in urn $A$, ${n_1} = 3$
Total number of balls in urn $B$, ${n_2} = 9$
The number of ways in which the two balls are selected from urn $A$ and from urn $B$ is the product of combination of urn $A$ and urn $B$.
$N = {n_1}{c_{{r_1}}} \times {n_2}{c_{{r_2}}}$
Where,
$N$ is the number of ways
$N = 2{c_2} \times 9{c_2}$
By using the formula of combination,
$N = \dfrac{{{n_1}!}}{{{r_1}!\left( {{n_1} - {r_1}} \right)!}} \times \dfrac{{{n_2}!}}{{{r_2}!\left( {{n_2} - {r_2}} \right)!}}$
Substituting the values of ${n_1}$, ${n_2}$, ${r_1}$ and ${r_2}$ in the above equation,
\[
  N = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{9!}}{{2!\left( {9 - 2} \right)!}} \\
  N = \dfrac{{3!}}{{2! \times 1!}} \times \dfrac{{9!}}{{2! \times 7!}} \\
 \]
By taking factorials in the above equation,
$N = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} \times \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}$
By cancelling the same terms in above equation,
$N = \dfrac{3}{1} \times \dfrac{{9 \times 8}}{2}$
By solving the above equation,
$
  N = 3 \times 9 \times 4 \\
  N = 108 \\
 $
The number of ways the balls are selected is, $N = 108$
Hence, by using $108$ ways the balls are selected from the urn $A$ and urn $B$.

Thus, the option (C) is correct.

Note: The difference between the combination and permutation is the consideration of order of selection. In permutation, order of choice is considered. Hence, the count by permutation is always greater than the combination. The theory of combination is used in the allocation of telephone numbers and mobile numbers. It is also used to find the population of the country. This helps in the counting in the more complex probability areas.