Permutations And Combinations Ques 8
- How many different nine-digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? $(2000,2 M)$
(a) 16
(b) 36
(c) 60
(d) 180
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Answer:
Correct Answer: 8.(c)
Solution:
- $X \quad X \quad X \quad X \quad X$. The four digits $3,3,5,5$ can be arranged at ( ) places in $\frac{4 !}{2 ! 2 !}=6$ ways.
The five digits $2,2,8,8,8$ can be arranged at $(X)$ places in $\frac{5 !}{2 ! 3 !}$ ways $=10$ ways.
Total number of arrangements $=6 \times 10=60$
[since, events $A$ and $B$ are independent, therefore $A \cap B=A \times B]$
BETA
