PandC aptitude tcs


#1

What will be the sum of all five digit numbers which can be formed with the digits 0, 1, 2,3,4 without repetition
A. 2599980
B. 235500
C. 923580
D. 765432


#2

If u observe each digit will be present at each position in 4! time.
i.e.,
No of times 0 is in first(ten thousand) place is (5-1)! = 4!= 4321 = 24
In the same way ,
Number of times 1 is in first place is (5-1)! = 4!= 4
321 = 24
Number of times 2 is in first place is (5-1)! = 4!= 4321 = 24
Number of times 3 is in first place is (5-1)! = 4!= 4
321 = 24
Number of times 4 is in first place is (5-1)! = 4!= 4321 = 24
Number of times 0 is in second( thousand) place is (5-1)! = 4!= 4
321 = 24
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.
.
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Number of times 4 in fifth(One’s) place is (5-1)! = 4!= 432*1 = 24

So the sum of all numbers will be
= (5-1)! * (10000+1000+100+10+1)(0+1+2+3+4)
= 4! * (11111)10
= 24
11111
10
= 2666640

Logic can be (n-1)!(111…1 ntimes)(sum of digits)

As numbers formed by digits 1,2,3,4 and 0 as first digits are not 5 digit numbers, so subtracting 4 digit numbers formed by 1,2,3,4
i.e., 2666640 - 3!(1111)(1+2+3+4)
= 2666640 - 6666*10
= 2666640 -66660
= 2599980