# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
 ✖ A. 420 ✖ B. 360 ✖ C. 320 ✔ D. 210

Solution:
Option(D) is correct

Total number of arrangements,

$= \dfrac{8!}{4!×2!×2!} = 420$

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 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 $= \textbf{210}$

## (2) Comment(s)

Numan
()

I think the answer is obtained by 7!/4!1!1! since you cant change the order of the first 1 and 2

Priyanka
()

why the 1st 1 before the first 2 is same as first 2 before the first 1