Question

How many arrangements of four $0^{\prime }s$ (zeroes), two $1^{\prime }s$ and two $2^{\prime }s$ are there in which the first $1$ occur before the first 2?

• A. $420$
• B. $360$
• C. $310$
• D. $210$

Answer 1

Answer: (D)

$210$

Number of zeros$=4$

Number of ones $=2$

Number of twos $=2$

Total numbers$=8$

Numerals to be arranged$=6$

Because remaining two are fixed in order

Total arrangements$=\frac{{}^8{P}_6}{2!2!4!}$

$=\frac{20160}{96}=210$

Hence the correct answer is $210$