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

Hint: 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 \[2\].