In how many ways can 6 boys and 2 girls be seated in a row if 2 girls must seat together? Expand (1 + x)^1/2 up to the term containing x^3 and use it to approximate √10.

Steps to Solve

1. Arrange Boys Together with Girls Since 2 girls must sit together, consider them as a single unit or block.

   • This means we now have 6 boys + 1 girl block = 7 units to arrange.
   • The number of ways to arrange these 7 units is given by $7!$.

2. Arrange Girls Inside Their Block Now, within the girl block, the 2 girls can be arranged among themselves.

   • The number of ways to arrange 2 girls is $2!$.

3. Calculate Total Arrangements Multiply the two results obtained above to find the total number of arrangements:

   $$ \text{Total Arrangements} = 7! \times 2! $$

   • Calculate $7! = 5040$ and $2! = 2$.
   • Thus, total arrangements = $5040 \times 2 = 10080$.

4. Expand the Series For the second part, we need to approximate $\sqrt{10}$ by expanding $(1 + x)^{\frac{1}{2}}$.

   • Rewrite $\sqrt{10}$ as $\sqrt{1 + 9}$. Let $x = 9$.
   • The binomial expansion is:

   $$(1 + x)^{\frac{1}{2}} = 1 + \frac{1}{2}x + \frac{1}{2} \left( \frac{1}{2} - 1 \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{1}{2} - 1 \right)\left( \frac{1}{2} - 2 \right) \frac{x^3}{3!} + \ldots$$

5. Evaluate Each Term for $x = 9$

   • Substitute $x = 9$ into each term of the expansion:

   $$ \sqrt{10} \approx 1 + \frac{1}{2}(9) + \frac{1}{2} \left( -\frac{1}{2} \right) \frac{9^2}{2} + \frac{1}{2} \left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right) \frac{9^3}{6} $$

6. Calculate Individual Terms Simplify each term:

   • ( 1 + \frac{9}{2} - \frac{9^2}{8} + \frac{27 \cdot 9^3}{48} )
   • Calculate each value step-by-step.