Induction Consider the set of bitstrings x ∈ {0, 1} n+k with n zeros and k ones with the additional condition that no ones are adjacent. (For n = 3 and k = 2, for example, the legal bitstrings are 00101, 01001, 01010, 10001, 10010, and 10100.) Prove by induction on n that the number of such bitstrings is n+1 k ? .

The answer is 6.

Step-by-step explanation:

From the question given,

Let us consider the set of bit strings x ∈ { 0.1} with n zeros and k ones.

The additional condition that no one are adjacent  for n=3 and k=2

Then

(n+1 k) = (n+1)ǃ/k ǃ (n+1 )ǃ = 4ǃ/2ǃ (2ǃ)

which is

n+1 k) = (n+1)ǃ/k ǃ (n+1 )ǃ = 4ǃ/2ǃ (2ǃ) = 24ǃ/ 2*2 =24/4

Therefore

24ǃ/ 2*2 =24/4 = 6

3. parallelogram

step-by-step explanation:

we turn each of our sides into direction vectors

ab=b-a=(3 - 7, 1 - 0) = (-4, 1)

bc=c-b=(1,6)

cd=d-c=(4,-1)

da=a-d=(-1,-6)

we see how the opposite sides have negated direction vectors; this is characteristic of a parallelogram.   we have congruent sides ab=cd and bc=da with unequal lengths because 1^2+6^2 is different from (-4)^2+1^2.   so it's not a square or a rhombus.  

we check the dot product of direction vectors of adjacent sides.   we have

ab dot bc = -4(1) + 1(6) = 2

that's not zero so the sides are not perpendicular.   we've ruled out rectangle.

the direction vectors indicate two pairs of congruent parallel sides, so we choose parallelogram over trapezoid.

whatever you can spend with $11.53 or under.

step-by-step explanation:

20 - 8.47 = 11.53. take away the money he used for lunch from the starting amount of 20.