A) \[\frac{1}{4}\]
B) \[-\frac{1}{4}\]
C) \[\frac{1}{8}\]
D) \[-\frac{1}{8}\]
Correct Answer: D
Solution :
There are 7 letters in the word FAILURE. To include F in every 4 letters word, we first select four letters from the 7 letters of the word FAILURE such that F is included in every selection. This can be done by selecting three letters from the remaining six letters. ie,\[A,\text{ }I,\text{ }L,\text{ }U,\text{ }R,\text{ }E\]in\[^{6}{{C}_{3}}\]ways. Now there are 4 letters in each of \[^{6}{{C}_{3}}\] selections. Also every selection contains letters which can be arranged in 4! ways. Hence, the required number of four letter words \[{{=}^{6}}{{C}_{3}}\times 4!=\frac{6.5.4}{3.2.1}.24\] \[=20\times 24=480\]
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