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

Concept:

The number of words from n letters where a letter is repeated m times

= n!/m!

where n! = n(n - 1)(n - 2)....3.2.1

Calculation:

If all the vowels come together, let us consider them as a single letter,

let α = (IEEEE)  {Vowels in the word INDEPENDENCE}

So, the number of words from the letters of the word INDEPENDENCE, in which vowels always come together

⇒ the number of words from the letters of the word NDPNDNC(IEEEE)

⇒ the number of words from the letters of the word NDPNDNCα = \(8!\over3!2!\) = 3360

{N is repeated 3 times and D is repeated two times}

And the total number of words that can be made by α 

{changing the vowels together}

= the number of words from letters IEEEE = 5!/4! = 5

So, the number of words that can be made out from the letters of the word INDEPENDENCE,

in which vowels always come together = 3360 × 5 = 16800.

∴ The correct answer is option (1).

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).

Concept:

• The number of ways in which r objects can be arranged in n places is ⁿP_r = \(\rm \dfrac{n!}{(n - r)!}\).
• The number of ways in which n objects, out of which p, q, r etc. are of same type, can be arranged is: \(\rm \dfrac{n!}{p!\ q!\ r!\ ...}\).
• The number of ways in which r distinct objects can be selected from a group of n distinct objects is ⁿC_r = \(\rm \dfrac{n!}{r!(n - r)!}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.
• If some objects cannot be together, then first arrange the remaining objects, and then arrange the objects which cannot be together in the empty spaces between them.

Calculation:

The word PROBABILITY has 11 letters out of which 4 (O, A, I, I) are vowels and 7 (P, R, B, B, L, T, Y) are consonants.

Since no two vowels can be together, we will first arrange the remaining 7 consonants.

The consonant can be arranged in \(\rm \dfrac{7!}{2!}=2520\) ways.

Now, the 4 vowels can be arranged in any of the 8 spaces between the consonants, as illustrated by a _ below:

_ P _ R _ B _ B _ L _ T _ Y _.

Number of ways of selecting some 4 places out of the 8 places = \(\rm ^8C_4 = \dfrac{8!}{4!(8 - 4)!}=70\).

Number of ways in which the 4 vowels can be arranged in these 4 places = \(\rm \dfrac{4!}{2!}=12\)

∴ The total number of possible arrangements = 2520 × 70 × 12 = 21,16,800.