[JEE Advanced 2000] How many nine-digit numbers from $223355888$ (by rearrangement) have the odd digits in even positions?
How many nine-digit numbers from $223355888$ (by rearrangement) have the odd digits in even positions?
(a) $16$
(b) $36$
(c) $60$
(d) $180$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (c) $60$
Odd digits $3,3,5,5$ in the $4$ even positions: $\frac{4!}{2!2!}=6$; even digits $2,2,8,8,8$ in the $5$ odd positions: $\frac{5!}{2!3!}=10$. Total $60$.
JEE Advanced 2000 · Permutations and Combinations — verified solution by the Vidaara Team.
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