Question: The total number of 5 digit numbers of different digits in which the digit in the middle is the largest, is
Options:
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
Show Answer
Answer:
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 with 0). Similarly, the number 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}} $
BETA
