Discussion
The number of eight-bit computer words formed by 0s and 1s with exactly three 1s and five 0s is :
Answer: B
There are 56 such words.
Choose which 3 of the 8 bit positions hold the 1s; the remaining 5 are 0s:
\[ \begin{aligned} \binom{8}{3} &= \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \\ &= 56 \end{aligned} \]
Choosing the positions of the five 0s gives the same answer, \(\binom{8}{5} = 56\). The value 15 is \(\binom{6}{2}\), and 30 and 65 do not arise from choosing positions at all.
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