NIMCET 2009 Question Paper with solutions

Question : 58 of 120

Marks: +1, -0

Using the digits

1,5,2,8

all possible four digit numbers are formed and the sum of all such numbers is between

Solution:

Sum of the numbers will be

(1+5+2+8) 3 !(1111)=96(1111)=106656

which fies hetween 100000 and 150000

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