| In how many ways can 10 books be arranged on a shelf, so that two specified books are not together? |
A) \[9!\,\,\times 2\]
B) \[9!\,\,\times 8\]
C) \[10!\,\,\times 2\]
D) \[9!\,\,\times 4\]
Correct Answer: B
Solution :
| Total number of arrangements of 10 books =10! If two specified books are always together, then the number of arrangements \[=\text{ }9!\text{ }\times \text{ }2!\] |
| Hence, the number of ways in which two specified books are not together\[=10!\,\,-9!\,\,\times 2!\] |
| \[=10\times 9!\,\,-2\times 9!=9!\,\,(8)\] |
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