A) \[\frac{17}{1944}\]
B) \[\frac{4}{27}\]
C) \[\frac{1}{945}\]
D) \[\frac{34}{243}\]
Correct Answer: C
Solution :
| Ten digits are split the digits into pairs (0, 1), (1, 2), (2, 3), ..., (8, 9) |
| Disjoints pairs out of these are (0, 1)(2, 3)(4, 5)(6, 7)(8, 9) |
| Now, two cases are |
| [a] When the pair (0,1) is not used for first and last place, the number of ways \[=4\times 4!\times {{2}^{5}}\] |
| [b] When the pair (0, 1) is used for first and last place, the number of ways \[=1\times 4!\times {{2}^{4}}\] |
| \[\therefore \]Total number of favourable eases \[=9\times 4!\times {{2}^{4}}\] |
| Required probability \[=\frac{9\times 4!\times {{2}^{4}}}{9\times 9!}=\frac{1}{945}\] |
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