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

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.

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!

Concept:

Permutation:

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

ⁿP_r 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+1P_q ways.

By the fundamental theorem,

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

 = \(\frac{p!.(p+1)!}{[(p+1)-q]!}\)

 = \(\frac{p!.(p+1)!}{[p-q+1]!}\)

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

Explanation:

We only need 2 points to form straight lines

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) = ⁿC_r 
• To arrange n things in an order of a number of objects taken r things = ⁿP_r

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:

ⁿP_r = \(\rm \frac{n!}{(n - r)!}\)

ⁿP_r = 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)
• \(\rm P_{k}^{n}\)
• _nP_k
• ⁿP_k 
• 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."

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, ..…, p_r are alike of rth kind such that p₁ + p₂ + ….+ p_r = 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.

Number of apartment = 3

Number of accepted student = 3 or 4

total number of student = 10

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

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.

Concept:

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

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).

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: 

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.