Question:

The sum of all the five digit numbers formed with the digits $1, 2, 3, 4, 5$ taken all at a time, is

Updated On: Jun 23, 2023
  • $15 (5!) $
  • $3999960$
  • $ 3990000 $
  • None of these
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The Correct Option is B

Solution and Explanation

Sum of unit's place $= 4!(1 + 2 + 3 + 4 + 5)$ $ = 24 \times 15 = 360$ [Since $1$ will come $24$ times in unit place and so the other digits if $5$ digit number is formed with digits $1, 2, 3, 4, 5 $ taken all at a time] So the sum in tens, hundreds, thousands and ten thousands places is $= 360 \times 10000 + 360 \times 1000 + 360 \times 100 + 360 \times 10 + 360$ $ = 3999960$.
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Concepts Used:

Arithmetic Progression

Arithmetic Progression (AP) is a mathematical series in which the difference between any two subsequent numbers is a fixed value.

For example, the natural number sequence 1, 2, 3, 4, 5, 6,... is an AP because the difference between two consecutive terms (say 1 and 2) is equal to one (2 -1). Even when dealing with odd and even numbers, the common difference between two consecutive words will be equal to 2.

In simpler words, an arithmetic progression is a collection of integers where each term is resulted by adding a fixed number to the preceding term apart from the first term.

For eg:- 4,6,8,10,12,14,16

We can notice Arithmetic Progression in our day-to-day lives too, for eg:- the number of days in a week, stacking chairs, etc.

Read More: Sum of First N Terms of an AP