The number of ways in which we can get at least 6 successive tails when a coin is tossed 10 times in a row, is

 The sequence of six tails may begin at with the ith toss where$i = 1, 2, 3, 4, 5.$Thus, the number of ways is

$2^4+1\left(2^3\right)+2\left(1\right)\left(2^2\right)+2^2\left(1\right)2+2^3\left(1\right)=48$