2 Marks Questions

Question 12 Marks

A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box, if atleast one black ball is to be included in the draw.
[Hint: Required number of ways =³C₁×⁶C₂+ ³C₂ × ⁶C₂ + ³C₃ .]

Answer

We have 2 white, 3 black and 4 red balls in a box. 3 balls are to be drawn out of 9 balls atleast one back ball is to be included So, the possible selection is (1 black and 2 other balls) or (2 black and 1 other ball) or (3 black and no other ball)
So, the number of possible selection is
= ³C₁ × ⁶C₂ + ³C₂ × ⁶C₁ + ³C₃ × ⁶C₀
=3 × 15 + 3 × 6 + 1 × 1 = 45 + 18 + 1 = 64
Hence, the required selection = 64.

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Question 22 Marks

Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
[Hint: 2 women occupy the chair, from 1 to 4 in ⁴P₂ ways and 3 men occupy the remaining chairs in ⁶P₃ ways.]

Answer

First the women choose the chairs fron amondst the chairs numberred 1 to 4.
Two woman can be arranged in 4 chairs in ⁴p₂ ways.
In remaining 6 chirs, 3 men can be arranged in ⁶p₃ ways.
$\therefore$ Total nimber of possible arrangements $=\ ^4\text{P}\times\ ^6\text{P}_3=\frac{4!}{2!}\times\frac{6!}{3!}$
$=4\times3\times6\times5\times4=1440$

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Question 32 Marks

How many committee of five persons with a chairperson can be selected from 12 persons.
[Hint: Chairman can be selected in 12 ways and remaining in 11C₄ .]

Answer

Total number of person to be selected = 5
Number of person to be selected = 5
Out of 5, there is a chairperson
$\therefore$ Number of ways of selecting a chairperson = ¹²C₁ = 12
Number of ways of selecting other 4 numbers out of remaining 11 persons = ¹¹C₄
$\therefore$ Total number of ways = ¹²C₁ × ¹¹C₄
$=12\times\frac{11.10.9.8}{4.3.2.1}=12\times330=3960$
Hence, the required number of ways = 3960

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Question 42 Marks

If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT?
[Hint: In each case number of words beginning with A, C, H, I is 5!]

Answer

The alphabetical order or ‘RACHIT is A, C, H, I, R and T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with I = 5!
And Number of word beginning with
$\therefore$ The rank of the word ‘RACHIT’ in the dictionary
= 5! + 5! + 5! + 5! + 1 = 4 × 5! + 1
= 4 × 5 × 4 × 3 × 2× 1 + 1 = 4 × 120 + 1 = 480 + 1 = 481

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Question 52 Marks

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

Answer

It is given that 2 questions are compulsory out of 5 questions.
So, the other 2 questions can be selected from the remaining 3 questions in ³C₂ = 3 ways.

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