- A$2^{7}-1$
- B$2^{8}-2$
- C$2^{8}-1$
- D$2^{8}$
Solution:
$({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{3}}+{^\text{7}}\text{C}_{\text{4}})+......$
$=1+2\times{^\text{7}}\text{C}_{\text{1}}+2\times{^\text{7}}\text{C}_{\text{2}}+2\times{^\text{7}}\text{C}_{\text{3}}+2\times{^\text{7}}\text{C}_{\text{4}}+2\times{^\text{7}}\text{C}_{\text{5}}..$
$=2+2^{2}({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})$
$=2+2^{2}(7+\frac{7}{2}\times6+\frac{7}{3}\times\frac{6}{2}\times5)$
$=2+252$
$=254$
$=2^{8}-2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Without repetition of the numbers, four digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is:
The total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}$ after simplification is:
$\lim\limits_{\text{x} \rightarrow0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}$ is equal to:
$\text{n}$
$1 $
$-\text{n}$
$0$