Question

How many arrangements of four $0's$( zeros) two $1's$ and two $2's$ are there in which the first $1$ occur before the first $2$?
A) $420$
B) $360$
C) $310$
D) $210$

Answer 1

In this type of problem we need to use concept of permutation. We know that a permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement, and use the formula for total arrangement.

Complete Step-by-step Solution
Step 1.
We know that,
Total arrangement is given by $\dfrac{{n!}}{{\left( {p! \times q! \times r!} \right)}}$
Hence,
Total number of arrangements,

$$= \dfrac{{8!}}{{4! \times 2! \times 2!}} \\\ = 420 \\\$$

Since there are two $1's$ and two $0's$, the number of arrangements in which the first $1$ is before the first $2$ is the same as the number of arrangement in which the first $2$ is before the first $1$ and they are each equal to half the total number of arrangements $= 210$.

$\therefore$ Option (D) is the correct option.

Note:
Half the total number of arrangements to find the number of arrangements of four $0's$ (zeros) two $1's$ and two $2's$ is there in which the first $1$ occur before the first $2$.