Answer to Question #186291 in Statistics and Probability for Aliyajaved

Question

a) A girl can take 3 books from her school library and there are 10 books of her interest in the library. Of these 10, she does not want to borrow Mechanics Part II, unless she borrows Mechanics Part I. In how many ways can she choose the 3 books to be borrowed? b) The number of ways in which letters of the word PARENT can be arranged so that R is always next to A is-------. c) From 4 boys and 8 girls, the number of ways can 6 be chosen, (i) to include only one boy. (ii) to include at least one boy. d) The number of numbers that can be formed with the digits 7, 6, 9, 8, 9, 6, 7 so that the odd digits always occupy the odd places ---------. e) A bag contains 2 white marbles, 3 black marbles and 4 red marbles. In how many ways can 3 marbles be drawn from the box if at least one black marble is to be included in the draw? 

Answer 1

(a) Number of ways she can borrow

When Mechanics Part I is borrowed = $^9C_2$

When Mechanics Part I is not borrowed = $^8C_3$

Total number of ways = $^9C_2+^8C_3$

$\Rightarrow \dfrac{9\times 8}{2}+\dfrac{8\times 7\times 6}{6}\\ \Rightarrow 36+56=92$

(b) There are 6 letters in the word PARENT

“You want R to appear right next to A”

Lets tie up R & A together so as they appear as one unit, i.e. AR.

Now, you have got 5 letters(or different items), i.e., P, AR, E, N, T, which can be arranged in 5! ways.

So answer should be 5! = 120.

(c) Total number of boys =4

total number of girls =8

Total members to be chosen = 6

(i) Number of ways so that committee contains only one boy = $^4C_1\times ^8C_5=4\times 56=224$

(ii) Number of ways so that committee contains only one boy =

$$^4C_1\cdot^8C_5+ ^4C_2\cdot ^8C_4+^4C_3\cdot ^8C_3+^4C_4\cdot ^8C_2$$

(d) We have been given seven digits, namely 7,6,9,8,9,6,7.

So, we have to form 7-digit numbers, so that odd digits occupy odd places.

Every 7 - digits number has 4 odd places.

Given four odd digits are 7,9,9,7 out of which 7 occurs 2 times and 9 occurs 2 times.

So, the number of ways to fill up 4 odd places = $\dfrac{4!}{2!\times 2!}=6$

Given three even digits are 6,8,6 in which 6 occurs 2 times and 8 occurs 1 time.

So, the number of ways to fill up 3 even places = $\dfrac{3!}{2!}=3$

Hence, the required number of numbers =(6×3)=18.

(e) According to question,

From 2 white marbles, 3 black marbles and 4 red marbles,.

3 marbles are to be selected such that at least one black ball should be there.

So, Number of choices were

All three are black $= \ ^3C_3$

Two are black and one is non black  $=\ ^3C_2\cdot ^6C_1$

One is black and two are non black $=\ ^3C_1\cdot ^6C_2$

Total number of ways are

$^3C_3+^3C_2\cdot^6C_1+\ ^3C_1\cdot ^6C_2\\=1+18+45\\=64$

Therefore, The number of ways are 64.