JEE PYQ: Permutation And Combination Question 29
Question 29 - 2020 (09 Jan Shift 1)
If the number of five digit numbers with distinct digits and 2 at the $10^{\text{th}}$ place is $336k$, then $k$ is equal to:
(1) 4
(2) 6
(3) 7
(4) 8
Show Answer
Answer: (4)
Solution
Five digit numbers with 2 at tens place: $__ __ 2 _$. First digit: 8 choices (1-9 except 2). Remaining 3 positions from 7 remaining digits: $8 \times 7 \times 6 = 336… $ Wait: $8 \times 8 \times 7 \times 6 = 2688$. Actually: tens digit fixed as 2. First digit: $1$-$9$ except $2$: 8 choices. Remaining 3 digits from 8 remaining: $8 \times 7 \times 6$. Total $= 8 \times 8 \times 7 \times 6 = 2688 = 336 \times 8$. So $k = 8$.
BETA
