Class 11th

7.

32. In the word DAUGHTER, there are 3 vowels, namely A, U, E and 5 consonants, namely D, G, H, T, R.

The number of ways of selecting 2 vowels out of 3

= ³C₂

= $\frac{3!}{2!\left(3-2\right)!}$

= 3

The number of ways of selecting 3 consonants out of 5

= ⁵C₃

= $\frac{5!}{3!\left(5-3\right)!}$

= 5 * 2

= 10

Therefore, the number of combination of 3 consonants and 2 vowels is 3 * 10 = 30.

Each of these 30 combinations has 5 letters which can be arranged among themselves in 5! Ways.

Therefore, the required numbers of different words is

= 30 * 5! = 30 * 5 * 4 * 3 * 2 * 1 = 3600

31. A student can choose 5 out of 9 available courses. However 2 specific courses are made compulsory.

So, now a student has 3 choices out of the remaining 7 courses.

Therefore, the required number of ways

=⁷C₃

= $\frac{7!}{3!\left(7?3\right)!}$

= 35

6. $\underset{x\to 0}{lim}\frac{{(x+1)}^5-1}{x}=\underset{x\to 0}{lim}\frac{x^5+5x^4+10x^3+10x^2+5x+1-1}{}$

Q6. $\underset{x\to 0}{\lim }\frac{{\left(x+1\right)}^5-1}{x}$

6. $\underset{x\to 0}{lim}\frac{{(x+1)}^5-1}{x}=\underset{x\to 0}{lim}\frac{x^5+5x^4+10x^3+10x^2+5x+1-1}{}$ $=\underset{x\to 0}{lim}{x}^4+5x^3+10x^2+10x+5.$ (0)4 + 5(0)3 + 10(0)2 + 10(0) + 5 = 5.

5. $\underset{x\to -1}{lim}\frac{x^{10}+x^5+1}{x-1}=\frac{{(-1)}^{10}+{(-1)}^5+1}{(-1)-1}=\frac{1-1+1}{-2}=-\frac{1}{2}$

4. $\underset{x\to 4}{lim}\frac{4x+3}{x-2}=\frac{4*4+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}$

3. $\underset{r\to 1}{lim}\pi r^2=\pi ·{(1)}^2=\pi$

30. We are given 5 black and 6 red balls of which 2 black and 3 red balls can be selected.

Thus the required number of ways

= ^5C₂*⁶C₃

= $\frac{5!}{2!\left(5-2\right)!}$ * $\frac{6!}{3!\left(6-3\right)!}$

= 10 * 20

= 200

2. $\underset{x\to \pi }{lim}\left(x-\frac{22}{7}\right)=\pi -\frac{22}{7}$

29. We are given 17 players of which 5 players can bowl and 17 – 5 = 12 can bat. But we need to select a team of 11 in which there are exactly 4 bowlers.

Hence, the required number of ways

=⁵C₄ (bowl) x ¹²C _(11-4) (bat)

= $\frac{5!}{4!\left(5-4\right)!}$ * $\frac{12!}{7!\left(12-7\right)!}$

= 5 * 11 * 9 * 8

= 3960