56
The number of triples of 1s on 8 bits
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Basically, this is the same as finding the number of distinct ways of arranging seven 1s and 3 0s. That is (10!/7!3!) = (10*9*8)/(3*2*1) = 120. There are 120 bit strings of length 10 with exactly three 0s.
It looks exactly like an Isosceles triangle, because two of its three sides have the same length.
The length of the longer leg of a right triangle is 3ftmore than three times the length of the shorter leg. The length of the hypotenuse is 4ftmore than three times the length of the shorter leg. Find the side lengths of the triangle.
A trinomial
it is...........two divided by three