Question

The number of $$6$$ digit numbers that can be formed using the digits $$0, 1, 2, 5, 7$$ and $$9$$ which are divisible by $$11$$ and no digit is repeated, is:

A
36
B
60
C
48
D
72

Solution

The correct option is C $$60$$Sum of given digits $$0, 1, 2, 5, 7, 9$$ is $$24$$.Let the six digit number be abcdef and to be divisible by $$11$$.So $$|(s + c + e) - (b + d + f)|$$ is multiple of $$11$$.Hence only possibility is $$a + c + e = 12 = b + d + f$$Case - I $$\{a, c, e\} = \{9, 2, 1\} \& \{b, d, f\} = \{7, 5, 0\}$$So, Number of numbers $$= 3! \times 3! = 36$$Case - II $$\{a, c, e\} = \{7, 5, 0\}$$ and $$\{b, d, f\} = \{9, 2, 1\}$$So, Number of numbers $$2 \times 2! \times 3! = 24$$Total $$= 60$$Mathematics

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