How many ways can the digits 2,3,4,5,6 be arranged to get a number divisible by 11
I know that the sum of the permutations of the digits should be divisible by 11. Also, the total number of ways the digits can be arranged is 5!=120.
Answer
Hint. By the divisibility rule by 11 we have to count the arrangements d1,d2,d3,d4,d5 of the digits 2,3,4,5,6 such that d1+d3+d5−(d2+d4) is divisible by 11. Notice that
−2=2+3+4−(5+6)≤d1+d3+d5−(d2+d4)≤4+5+6−(2+3)=10
therefore we should have d1+d3+d5=d2+d4=2+3+4+5+62=10.
In how many ways we can do that?
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