Permutation and CombinationHard
Question
The sum of all the numbers which can be formed by using the digits 1, 3, 5, 7 all at a time and which have no digit repeated, is
Options
A.16 × 4!
B.1111 × 3!
C.16 × 1111 × 3!
D.16 × 1111 × 4!
Solution
If 1 be unit digit then total no. of number is 3! = 6
Similarly so on if 3, 5, or 7 be unit digit number
then total no. of no. is 3! = 6
Hence sum of all unit digit no. is = 6 × (1+3+5+7)
= 6 × 16 = 96
Hence total sum is
= 96 × 103 + 96 × 102 + 96 × 101 + 96 × 100
= 96000 + 9600 + 960 + 96 = 106656
= 16 × 1111 × 3!
Similarly so on if 3, 5, or 7 be unit digit number
then total no. of no. is 3! = 6
Hence sum of all unit digit no. is = 6 × (1+3+5+7)
= 6 × 16 = 96
Hence total sum is
= 96 × 103 + 96 × 102 + 96 × 101 + 96 × 100
= 96000 + 9600 + 960 + 96 = 106656
= 16 × 1111 × 3!
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