All possible numbers are formed using the digits $1, 1, 2, 2, 2, 2, 3, 4, 4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is

• A

  $175$
• B

  $162$
• Correct answer
  C

  $180$
• D

  $160$

Solution

Odd digits: $1, 1, 3$
Even digits: $2, 2, 2, 2, 4, 4$
The number of ways of placing odd digits at even position $= {}^{4}C_3\times \frac{3!}{2!}=4\times 3=12$
The number of ways of placing even digits $=\frac{6!}{4! 2!}=15$
$\therefore$ Total number of ways $=12\times 15=180$

Hence, the correct answer is option: 'C'.