How many six digit no can b formed using the digits 1 2 3 4 5 6 each exactly once such tht they are multiples of 11
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6 can be only at one of the last three places 3 possibility
6 is 5 each also come only in one of last 3 remaining places.
Continuing similarly we get.
Total no. of possibilities are 3
3
×2=162
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