True False[1 Marks ]

Question 11 Mark

In a steamer there are stalls for $12$ animals, and there are horses, cows and calves $($not less than $12$ each$)$ ready to be shipped. They can be loaded in $3^{12}$ ways

Answer

True.
In each stall any one of the three animals can be shipped. So total number of ways of loading $= 3 \times 3 \times 3 \times … \times 12\ \times = 3^{12}$

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Question 21 Mark

Eighteen guests are to be seated$,$ half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table.
The number of ways in which the seating arrangements can be made is $\frac{11!}{5!6!}(9!)(9!) .$

Answer

True.
Let two sides of the table be $A$ ans $B$ each having $9$ seats.
Let on side $A,$ four particular guests and on side $B,$ three particular seated.
Now for side $ A,$ five more guests can be selected from remaining eleven guests in $^{11}C_5$ ways.
Also on each side nine guests can be arranged in $9!$ ways.
So total number of ways of arrangement $=\ ^{11}\text{C}_5\times\ 9!\times9!=\frac{11!}{5!\ 6!}(9!)(9!)$

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Question 31 Mark

There are $12$ points in a plane of which $5$ points are collinear, then the number of lines obtained by joining these points in pairs is $^{12}C_2 – ^5C_2 .$

Answer

False.
Required number of lines $= \ ^{12}C_2 – ^5C_2 + 1$ Hence$,$ the given statement is 'False'

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Question 41 Mark

To fill $12$ vacancies there are $25$ candidates of which$ 5$ are from scheduled castes. If $3$ of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is $\ ^5C_3 \times \ ^{20}C_9 .$
In each if the Exercises from $60$ to$ 64$ match each item given under the column $C_1$ to its correct answer given under the column $C_2 $.

Answer

False.
Number of whay to select $3$ scheduled caste candidates out of $5 = \ ^5C_3$
We have to select $9$ other candidates out of $22.$
So$,$ the no of whay $= \ ^{22}C_9$
Required number of selection $= \ ^5C_3 \times \ ^{22}C_9$
Hence$,$ the given statement is false.

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Question 51 Mark

Three letters can be posted in five letterboxes in $35$ ways.

Answer

False.
Given that $3$ letters are to be posted in $5$ letter boxes.
$\therefore$ Required number of ways $= 5^3 = 125$
Hence$,$ the given statements is ‘False’.

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Question 61 Mark

If some or all of $n$ objects are taken at a time$,$ the number of combinations is $2^{n–1}.$

Answer

True.
When some or all objects$,$ taken at a time$,$ then the number of selection will be$,$
$\ ^nC_1 + \ ^nC_2 + \ ^nC_3 + … + \ ^nC_n = 2^{n – 1}$
$\because \ ^nC_0 + \ ^nC_1 + \ ^nC_2 + … + \ ^nC_n = 2n$
Hence$,$ the given statement is ‘True’.

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Question 71 Mark

In the permutations of $n$ things$,$ r taken together$,$ the number of permutations in which $m$ particular things occur together is $ ^{n–m}P_{r–m} \times ^rP_m .$

Answer

False.
Arrangement of $n$ things$,$ taken $r$ at a time in which $m$ things occur together first we select $(r - m)$ objects $(n - m)$ object in $^{n - m}C_{r - m}$ ways.
Now we consider there m things as $1$ group.
Number of objects excluding these $m$ objects $= (r - m)$
Now fast we have to arrange $(r - m + 1)$ objects.
Number of arrengement $= (r - m + 1)!$
Also m objects which we considered as $1$ geoup$,$ can be arranged in $m!$ ways.
$\therefore$ Required number of arrangement $= \ ^{n - m}C_{r - m} \times (r - m + 1)! \times m!$

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Question 81 Mark

There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.

Answer

False
Solution:
Number of ways of selecting any number of objects from given n identical objects is 1.
Now selecting zero or more red ball from 4 identical red balls = 1 + 1 + 1 + 1 + 1 = 5
Selecting at least 1 black ball from 5 identical black balls =1 + T + 1 + 1 + 1= 5 So, total number of ways = 5 × 5 = 25

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Question 91 Mark

A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.

Answer

False.
Solution:
A candidate can attempt questions in following maimer

Group A	2	3	4	5
Group B	5	4	3	2

So, the Number of attempts of 7 question
$=\ ^6\text{C}_2\times\ ^6\text{C}_5+\ 6\text{C}_3\times\ ^6\text{C}_4+\ ^6\text{C}_4+\ ^6\text{C}_3+\ ^6\text{C}_5\times\ ^6\text{C}_2$
$=2[\ ^6\text{C}_2\times\ ^6\text{C}_5+\ ^6\text{C}_3\times\ ^6\text{C}_4$
$=2[15\times6+20\times15]=2[90+300]=2\times390=780$

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