ankitranjan wrote:
Find the sum of all the four digit numbers formed using the digits 2,3,4 and 5 without repetition.
I will post OA and OE tomorrow.
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Along with the question, do post the options.
if you keep the digit 2 at the one's digit. The total numbers that can be formed is 3! = 6
=> 2 occurs at one's digit 6 times. Similarly all the 4 numbers occurs at one's digit 6 times.
Similar all the 4 numbers occurs at all the 4 digits - one's,ten's, hundred's, thousand's- 6 times.
=> total sum = ( 1000+100+10+1 ) *( 2+3+4+5)*6
= 1111*14*6 = 1111*84 = 93324
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