Permutation-Combination
Aptitude

 Back to Questions
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

 Hide Ans

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