JEE PYQ: Permutation And Combination Question 31
Question 31 - 2019 (08 Apr Shift 1)
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:
(1) 180
(2) 175
(3) 160
(4) 162
Show Answer
Answer: (1)
Solution
9 digits total. Even places: 2nd, 4th, 6th, 8th (4 places). Odd digits: 1, 1, 3 (3 digits). Place 3 odd digits in 4 even places: ${}^4C_3 \times \frac{3!}{2!} = 4 \times 3 = 12$. Remaining 6 digits (2,2,2,2,4,4) in 6 places: $\frac{6!}{4!2!} = 15$. Total $= 12 \times 15 = 180$.
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