A) \[\sum\limits_{n=4}^{9}{^{n}{{P}_{4}}}\]
B) \[\sum\limits_{n=4}^{9}{^{n}{{P}_{4}}}-\frac{1}{3!}\sum\limits_{n=3}^{9}{^{n}{{P}_{3}}}\]
C) \[30(3!)\]
D) None of these
Correct Answer: D
Solution :
[d] Since the largest digit is in the middle, the middle digit is greater than or equal to 4, the number of numbers with 4 in the middle \[{{=}^{4}}{{P}_{4}}{{-}^{3}}{{P}_{3}}.\] (\[\because \] The other four places are to be filled by 0, 1, 2 and 3, and a number cannot begin with0). Similarly, the numbers of numbers with 5 in the middle \[{{=}^{5}}{{P}_{4}}{{-}^{4}}{{P}_{3}},\] etc.) \[\therefore \] The required number of numbers \[={{(}^{4}}{{P}_{4}}{{-}^{3}}{{P}_{3}})+{{(}^{5}}{{P}_{4}}{{-}^{4}}{{P}_{3}})+{{(}^{6}}{{P}_{4}}{{-}^{5}}{{P}_{3}})+...+{{(}^{9}}{{P}_{4}}{{-}^{8}}{{P}_{3}})\]\[=\sum\limits_{n=4}^{9}{^{n}{{P}_{4}}-\sum\limits_{n=3}^{8}{^{n}{{P}_{3}}}}\]
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