2 Marks Questions

Question 12 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 $11C_4 .]$

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 $= ^{12}C_1 = 12$
Number of ways of selecting other $4$ numbers out of remaining $11$ persons $= ^{11}C_4$
$\therefore$ Total number of ways $= ^{12}C_1 \times ^{11}C_4$
$=12\times\frac{11.10.9.8}{4.3.2.1}=12\times330=3960$
Hence$,$ the required number of ways $= 3960$

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Question 22 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 $ =\ ^3C_1\times \ ^6C_2+ \ ^3C_2 \times \ ^6C_2 + \ ^3C_3 .]$

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
$= \ ^3C_1 \times \ ^6C_2 + \ ^3C_2 \times \ ^6C_1 + \ ^3C_3 \times \ ^6C_0$
$=3 \times 15 + 3 \times 6 + 1 \times 1 = 45 + 18 + 1 = 64$
Hence$,$ the required selection $= 64.$

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Question 32 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 $^4P_2$ ways and $3$ men occupy the remaining chairs in $^6P_3$ 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 $^4p_2$ ways.
In remaining $6$ chirs$, 3$ men can be arranged in $^6p_3$ 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 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 $^3C_2 = 3$ ways.

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