True False[1 Marks ]

Question 11 Mark

In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is ^n–mP_r–m × ^rP_m .

Answer

False.
Solution:
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} × (r - m + 1)! × m!

View full question & answer→

Question 21 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

View full question & answer→

Question 31 Mark

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

Answer

False.
Solution:
Given that 3 letters are to be posted in 5 letter boxes.
$\therefore$ Required number of ways = 5³ = 125
Hence, the given statements is ‘False’.

View full question & answer→

Question 41 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.
Solution:
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 ¹¹C₅ ways.
Also on each side nine guests can be arranged in 9! ways.
So total number of ways of arrangement $=\ ^{11}\text{C}_5\times\ 9!\times9!=\frac{11!}{5!\ 6!}(9!)(9!)$

View full question & answer→

Question 51 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 ¹²C₂ – ⁵C₂ .

Answer

False.
Solution:
Required number of lines = ¹²C₂ – ⁵C₂ + 1 Hence, the given statement is ‘False’

View full question & answer→

Question 61 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 ⁵C₃ × ²⁰C₉ .
In each if the Exercises from 60 to 64 match each item given under the column C1 to its correct answer given under the column C₂ .

Answer

False.
Solution:
Number of whay to select 3 scheduled caste candidates out of 5 = ⁵C₃
We have to select 9 other candidates out of 22.
So, the no of whay = ²²C₉
Required number of selection = ⁵C₃ × ²²C₉
Hence, the given statement is false.

View full question & answer→

Question 71 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
$=\ ^6\text{C}_2\times\ ^6\text{C}_5+\ 6\text{C}_3\times\ ^6\text{C}_4+\ ^6\text{C}_4+\ ^6\text{C}_3+\ ^6\text{C}_5\times\ ^6\text{C}_2$
$=2[\ ^6\text{C}_2\times\ ^6\text{C}_5+\ ^6\text{C}_3\times\ ^6\text{C}_4$
$=2[15\times6+20\times15]=2[90+300]=2\times390=780$

View full question & answer→

Question 81 Mark

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

Answer

True.
Solution:
When some or all objects, taken at a time, then the number of selection will be,
ⁿC₁ + ⁿC₂ + ⁿC₃ + … + ⁿC_n = 2^{n – 1}[$\because$ ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿC_n = 2n]
Hence, the given statement is ‘True’.

View full question & answer→

Question 91 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 3¹² ways

Answer

True.
Solution:
In each stall any one of the three animals can be shipped. So total number of ways of loading = 3 × 3 × 3 × … × l2 times = 3¹²

View full question & answer→

MATHS True False[1 Marks ]

Test yourself on this topic