• # question_answer The total number of 5 digit numbers of different digits in which the digit in the middle is the largest, is 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

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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