Word Problems MCQ Quiz in தமிழ் - Objective Question with Answer for Word Problems - இலவச PDF ஐப் பதிவிறக்கவும்

Last updated on Mar 21, 2025

பெறு Word Problems பதில்கள் மற்றும் விரிவான தீர்வுகளுடன் கூடிய பல தேர்வு கேள்விகள் (MCQ வினாடிவினா). இவற்றை இலவசமாகப் பதிவிறக்கவும் Word Problems MCQ வினாடி வினா Pdf மற்றும் வங்கி, SSC, ரயில்வே, UPSC, மாநில PSC போன்ற உங்களின் வரவிருக்கும் தேர்வுகளுக்குத் தயாராகுங்கள்.

Latest Word Problems MCQ Objective Questions

Top Word Problems MCQ Objective Questions

Word Problems Question 1:

60 is the total number of words with or without definition which can be made using all letters of the word AGAIN. Figure-out the 49th word if these words are mentioned as in wordbook?

  1. NAGIA
  2. NAIGA
  3. NAAIG
  4. NAAGI

Answer (Detailed Solution Below)

Option 4 : NAAGI

Word Problems Question 1 Detailed Solution

Given:

60 is the total number of words with or without definition which can be made using all letters of the word AGAIN.

Calculation:

While arranging these words in the dictionary, the words beginning with A will come first.

Fixing the first letter like A, we arrange the remaining 4 letters taken all at a time.

So, the number of words starting with A  = 4P4 = 4! = 24

Fix G in the first place and the remaining 4 letters in which no of A are 2 

The number of words starting with G = 4!/2! = 12

The number of words starting with I = 4!/2! = 12

Total number of words so far obtained = 24 + 12 + 12 = 48

Hence, the 49th word will start from N and the remaining four rearranged according to the dictionary.

The 49th word is NAAGI.

∴ The 49th word is NAAGI.

Word Problems Question 2:

Find the total number of 9 digit number which have all different digits:

  1. 9!
  2. 9 × 9!
  3. 8!
  4. 8 × 9!

Answer (Detailed Solution Below)

Option 2 : 9 × 9!

Word Problems Question 2 Detailed Solution

Concept:

PERMUTATION: Permutation:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

Number of Permutations of ‘n’ objects where there are n1 repeated items, n2 repeated items, ….. nk repeated items taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{{{\rm{n}}_1}!{{\rm{n}}_2}!{{\rm{n}}_3}! \ldots .{{\rm{n}}_{\rm{k}}}!}}\)

\(^np_0\ =\ 1\ and\ \ ^np_n\ =\ n!\)

Calculation:

A total number of 9 digit numbers having different digit is given by :

\({^{10}P_9} - {^9P_8} = \frac{10!}{1!}- \frac{9!}{1!}= 10! - 9!\)

= 10 × 9! - 9!

= 9! (10 - 1)

= 9! × 9

Hence, option 2 is correct.

Alternate MethodWe know that the total number of digits = 10 i.e.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Out of which 0 can't be placed in the first place. 

Therefore, first place can be filled in 9 ways.

The rest 9 blanks with 9 digits in 9! ways 

Hence, a total number of ways = 9 × 9!

Word Problems Question 3:

p men and q women are to be seated in a row so that no two women sit together. If p > q, then the number of ways in which they can be seated is

  1. \(\frac{p!.(p+1)!}{[p-q+1]!}\)
  2. \(\frac{p!.(p-1)!}{[p-q+1]!}\)
  3. \(\frac{(p-1)!.(p+1)!}{[p-q+1]!}\)
  4. \(\frac{p!.(p+1)!}{[p+q+1]!}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{p!.(p+1)!}{[p-q+1]!}\)

Word Problems Question 3 Detailed Solution

Concept:

Permutation:

Permutation is an arrangement of objects in a particular way or order.

nPr represents the “n” objects to be selected from “r” objects without repetition, in which the order matters.

Calculation:

Given: q < p 

Since p is more in number than q so first arrange the man first and in between the make if the man, q women will be arranged for the desired result.

Arrangement of p men, in a row = p! ways.

Since, no two women can sit together, in any one of the p! arrangements, there are (p + 1) places in which q women can be arranged in p+1Pq ways.

By the fundamental theorem,

The required number of arrangements of p men and n women(q < p) = p! p+1Pq

 = \(\frac{p!.(p+1)!}{[(p+1)-q]!}\)

 = \(\frac{p!.(p+1)!}{[p-q+1]!}\)

∴ The number of ways is \(\frac{p!.(p+1)!}{[p-q+1]!}\).

Word Problems Question 4:

Suppose 20 distinct points are placed randomly on a circle. Which of the following statements is/are correct?

1. The number of straight lines that can be drawn by joining any two of these points is 380.

2. The number of triangles that can be drawn by joining any three of these points is 1140.

Select the correct answer using the code given below.

  1. 1 only
  2. 2 only
  3. Both 1 and 2
  4. Neither 1 nor 2

Answer (Detailed Solution Below)

Option 2 : 2 only

Word Problems Question 4 Detailed Solution

Explanation:

We only need 2 points to form straight lines

We have 20 points so the number of straight lines drawn is 20C​= 190
 

We only need 3 points to form a triangle

We have 20 points, so number of triangles drawn is 20C3 = 1140

∴ Statement 1 is incorrect and Statement 2 is correct.

Word Problems Question 5:

How many different permutations can be made out of the letters of the word 'INDIANARMY' ?

  1. \(\rm \frac{8!}{(2!)^2}\)
  2. \(\rm \frac{8!}{(2!)^3}\)
  3. \(\rm \frac{10!}{(2!)2}\)
  4. \(\rm 10!\over{(2!)^3}\)

Answer (Detailed Solution Below)

Option 4 : \(\rm 10!\over{(2!)^3}\)

Word Problems Question 5 Detailed Solution

Concept:

  • The ways of arranging n different things = n!
  • The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
  • The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
  • The number of ways for selecting r from a group of n (n > r) = nCr 
  • To arrange n things in an order of a number of objects taken r things = nPr


Calculation:

The total number of words in INDIANARMY is 10

Here A, I and N are repeating twice 

So, Number of different permutations = \(\rm \frac{10!}{2!\ \times\ 2!\ \times\ 2!}\)

\(\rm 10!\over{(2!)^3}\)

Additional Information

Permutation: Permutation is a way of changing or arranging the elements or objects in a linear order.

The number of permutations of 'n' objects taken 'r' at a time is determined by the following formula:

nPr = \(\rm \frac{n!}{(n - r)!}\)

nPr = permutation

n = total number of objects
r = number of objects selected

The factorial function (Symbol: !) just means to multiply a series of descending natural numbers.

For examples:

4! = 4 × 3 × 2 × 1 

1! = 1

There are three types of permutation:

  1. Permutations with Repetition
  2. Permutations without Repetition
  3. Permutation when the objects are not distinct (Permutation of multi-sets)

Representation of Permutation:

We can represent in many ways such as: 

  • P (n, k)
  • \(\rm P_{k}^{n}\)
  • nPk
  • nPk 
  • P n, k

Application of Permutations:

  • Permutations are important in a variety of counting problems (particularly those in which order is important).
  • Permutations are used to define the determinant.

Important Points

Order is very important in permutation.

"A Permutation is an ordered combination."

Permutation Combination
Permutation means the selection of objects, where the order of selection matters The combination means the selection of objects, in which the order of selection does not matter.
In other words, it is the arrangement of r objects taken out of n objects.  In other words, it is the selection of r objects taken out of n objects irrespective of the object arrangement.
The formula for permutation is nPr = \(\rm \frac{n!}{(n - r)!}\)

The formula for combination is 

nCr = \(\rm \frac{n!}{r!(n - r)!}\)

Word Problems Question 6:

Find the number of arrangements of the letters of the word, INDEPENDENCE

  1. 1660000
  2. 1326600
  3. 1663200
  4. 1324600

Answer (Detailed Solution Below)

Option 3 : 1663200

Word Problems Question 6 Detailed Solution

Concept:

Let there be n things of which p1 are alike of one kind, p2 are alike of another kind, p3 are alike of 3rd kind, ..…, pr are alike of rth kind such that p1 + p2 + ….+ pr = n. Then the permutations of n objects is  \(\rm \frac{n!}{(p_1!)\times (p_2!)\times ....\times (p_r!)} \)

 

Calculation:

The word INDEPENDENCE contains 12 letters out of which N occurs 3 times, D occurs 2 times, E occurs 4 times, and the rest of the letters occur only once.

Hence, required number of arrangements

\(=\rm \frac{12!}{(3!)\times (4!)\times(2!)}\)

\(=\rm \frac{12\times 11\times10\times9\times8\times7\times6\times5\times4!​​}{(3\times2\times1)\times (4!)\times(2)}\)

\(=\rm 11\times10\times9\times8\times7\times6\times5​​\)

= 1663200

Hence, option (3) is correct.

Word Problems Question 7:

A building has 3 apartments A, B and C for rent. Each apartment accepts 3 or 4 residents. Find the number of ways in which apartments can be given to 10 students?

  1. 15000
  2. 12600
  3. 13500
  4. 10800

Answer (Detailed Solution Below)

Option 2 : 12600

Word Problems Question 7 Detailed Solution

Number of apartment = 3

Number of accepted student = 3 or 4

total number of student = 10

A B C
3 3 4
3 4 3
4 3 3

for apartment A, number of ways

\({10_{{C_3}}} × {7_{{C_3}}} × {4_{{C_4}}}\)

\(\frac{{10!}}{{7! × 3!}} × \frac{{7!}}{{3! × 4!}} × \frac{{4!}}{{4! × 10!}}\)

\(\frac{{10!}}{{3! × 3! × 4!}} = \)\(\frac{{10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1}}{{6 × 6 × 24}}\)

= 4200

Now there are such 3 possible arrangement thus, 

Total arrangements = 3 × 4200 = 12600

Word Problems Question 8:

A linguistic club of a certain Institute consists of 6 girls and 4 boys. A team of 4 members to be selected from this group including the selection of a Captain (from among these 4 members) for the team. If the team has to include atmost one boy, the number of ways of selecting the team is

  1. 95
  2. 260
  3. 320
  4. 380

Answer (Detailed Solution Below)

Option 4 : 380

Word Problems Question 8 Detailed Solution

Calculation:

Case I : No boy is included.

Selecting 4 girls from 6 girls = 6C4

Selecting 1 captain from selected members = 4C1

Total number of ways = 6C4 × 4C1 = 60

Case II : One boy is included.

Selecting 3 girls and 1 boy from given members = 6C3 × 4C1.

Selecting 1 captain from the selected members = 4C1.

Total Number of ways = 6C3 × 4C1 × 4C1 = 320.

∴ Total Number of ways = 320 + 60 = 380.

The correct answer is Option 4.

Word Problems Question 9:

Comprehension:

Consider the word 'QUESTION' :

How many 4-letter words each of two vowels and two consonants with or without meaning, can be formed ?

  1. 36
  2. 144
  3. 576
  4. 864

Answer (Detailed Solution Below)

Option 4 : 864

Word Problems Question 9 Detailed Solution

Concept:

The number of ways of choosing m objects out of n different objects = nCm

The number of ways of arranging n objects = n!

Explanation:

Given word 'QUESTION' which has 8 different letters

{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}

The number of ways of choosing two vowels from 4 vowels = 4C2 = 6

And the number of ways of choosing two vowels from 4 vowels = 4C2 = 6

⇒ The number of ways of choosing two vowels and two consonants from 4 vowels and 4 consonants = 4C2 × 4C2 = 36

And the number of ways of arranging these four different letters = 4! = 24

⇒ Number of 4-letter words each of two vowels and two consonants with or without meaning = 36 × 24 = 864

∴ The correct option is (4).

Word Problems Question 10:

The sides AB, BC, CA of a Δ ABC have 3, 5 and 6 interior points, respectively on them. How many triangles can be made by using these points as vertices?

  1. 364
  2. 333
  3. 240
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 333

Word Problems Question 10 Detailed Solution

Given :

AB, BC, CA are the sides of ΔABC.

AB, BC and CA have 3, 5 and 6 interior points.

Formula used : 

\(^nC_r = \frac{n!}{r!(n-r)!}\)       ----- (1)

Calculations: 

F1 Amar T 08-2-22 Savita D3

Total number of points on the sides of ΔABC = 3 + 5 + 6 = 14

The number of ways to form a triangle = \(^{14}C_3\)         ------- (2)

But we cannot select the 3 points from the same side.

⇒ \(^3C_3 + ^5C_3 + ^6C_3\)       ------ (3)

Subtracting equation (3) from (2), we get 

⇒ \(^{14}C_3 - (^{3}C_3 + ^{5}C_3 + ^6C_3)\)

Using equation (1)

⇒ \(\frac{14\times 13\times 12}{1\times 2\times 3} - (1+\frac{5\times 4}{1\times 2} + \frac{6\times 5\times 4}{1\times 2\times 3})\)

⇒ 364 - 31

⇒ 333

∴ The number of triangles that can be constructed using these points as vertices is 333.

Get Free Access Now
Hot Links: teen patti master real cash teen patti joy 51 bonus online teen patti