Question

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

• A. $\sum_{n = 4}^{9}{nP_{4}}$
• B. $\sum_{n = 4}^{9}{nP_{4}}$– $\frac { 1 } { 3 ! }$ $\sum_{n = 3}^{9}{nP_{3}}$
• C. 30 (3!)
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

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 = ⁴P₄ – ³P₃.

(Q 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 = ⁵P₄ – ⁴P₃, etc.

\ The required number of numbers

= (⁴P₄ – ³P₃) + (⁵P₄ – ⁴P₃) + (⁶P₄ – ⁵P₃) +…. + (⁹P₄ – ⁸P₃)

= $\sum_{n = 4}^{9}{nP_{4} - \sum_{n = 3}^{8}{nP_{3}}}$.