A) \[7!\]
B) \[2\times 7!\]
C) \[3\times 7!\]
D) \[4\times 7!\]
Correct Answer: D
Solution :
[d] Sum of 7 digits = a multiple of 9 Since sum of number 1, 2, 3? 8, 9 is 45 (Since a number is divisible by 9 if sum of its digits is divisible by 9.) So, two left number should also have sum as 9. The pairs to be left are (1, 8) (2, 7) (3, 6) (4, 5) with each pair left, number of 7 digit numbers\[=7!\]. So, with all 4 pair?s total seven digits number \[=4\times 7!\]
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