Question

There are 12 different balls in an urn. Number of ways in which 3 balls can be drawn from the urn four times in succession without replacement is equal to.
A. $\dfrac{12!}{{{\left( 3! \right)}^{4}}4!}$
B. $\dfrac{12!}{{{\left( 3! \right)}^{4}}}$
C. $\dfrac{12!}{{{9}^{2}},{{4}^{4}}}$
D. $\dfrac{12!}{{{\left( 4! \right)}^{3}}}$

Answer 1

Hint: We will find the ways in which 3 balls can be drawn from the urn each time. First we draw 3 balls from the urn which has 12 balls and without replacement we draw another 3 balls from the urn which now has $12-3=9$ balls. For that the ways will be ${}^{9}{{C}_{3}}$. In this manner, we will withdraw all 12 balls four times. Total ways will be multiplication of ways in each time divide by 4! (it is for selecting the withdraw 1 case).

Complete step by step solution:
There are a total 12 balls in the urn. And we have to withdraw 3 balls in ${{1}^{st}}$ withdraw.
So, number of ways in which ${{1}^{st}}$ withdraw can be drawn are,
$={}^{12}{{C}_{3}}............\left( 1 \right)$ (Total balls are 12 in which we have to select 3 balls)
Now, remaining balls (because without replacement is there) $=12-3=9$ balls.
Now, the ways in which ${{2}^{nd}}$ withdraw can be drawn are,
$={}^{9}{{C}_{3}}............\left( 2 \right)$
Remaining balls in urn $=9-3=6$.
Number of ways in which ${{3}^{rd}}$ withdraw can be drawn are,
$={}^{6}{{C}_{3}}............\left( 3 \right)$
Now, for the last withdrawal number of remaining balls in urn $=6-3=3$.
So, the ways in which ${{4}^{th}}$ withdrawal can be drawn are,
$={}^{3}{{C}_{3}}............\left( 4 \right)$
Now, total ways in 3 balls can be drawn from the urn four times in succession without replacement.
$=\dfrac{\text{Multipling the number of ways in each time}}{4!}.........\left( 5 \right)$
(4! is used for selecting the withdrawal cases as there are a total 4 withdraw cases).
\[={}^{12}{{C}_{3}}\times {}^{9}{{C}_{3}}\times {}^{6}{{C}_{3}}\times {}^{3}{{C}_{3}}\times \dfrac{1}{4!}\]
And we know that,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Applying the formula,
\[\begin{align}
  & =\dfrac{12!}{3!\times 9!}\times \dfrac{9!}{3!6!}\times \dfrac{6!}{3!3!}\times \dfrac{3!}{0!3!}\times \dfrac{1}{4!} \\
 & =\dfrac{12!}{3!}\times \dfrac{1}{3!}\times \dfrac{1}{3!3!}\times \dfrac{1}{4!} \\
 & =\dfrac{12!}{{{\left( 3! \right)}^{4}}4!} \\
\end{align}\]
So, the total number of ways are \[\dfrac{12!}{{{\left( 3! \right)}^{4}}4!}\].
Hence, the correct option is (A).

Note: Students may have the possibility of making a mistake that they may not divide by 4! In formula given in equation (5) which leads to the wrong answer as $\dfrac{12!}{{{\left( 3! \right)}^{4}}}$. Then they may choose option B as the right answer. Each time 3 balls are drawn without replacement, so in each of the four draws, we have to deduct 3 from the corresponding number of remaining balls. Some students may just deduct this once.