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Correct option is (b) Multiplication of 1 to n natural numbers

In a pack of 52 cards, there are (13 spade + 13 club) 26 cards of black colour and (13 heart + 13 diamond) 26 cards of red colour.

∴ From a well shuffled pack of 52 cards, 3 cards of same colour can be selected

In ²⁶C₃ + ²⁶C₃ ways.

∴ Total combinations = ²⁶C₃ + ²⁶C₃

= \(\frac{26×25×24}{3×2×1}+\frac{26×25×24}{3×2×1}\)

= 2600 + 2500 = 5200

First five natural numbers are 1 2 3 4 5

(i) Total numbers can be formed = ⁵P₅

= 5!

= 120

(ii) For the numbers greater than 30,000, the digit at the first place can by any one of the digits 3, 4, 5 and remaining 4 places can be arranged from the remaining 4 digits.

.-. Total permutations = ³P₁ × ⁴P₄

= 3 × 4!

= 3 × 24

= 72

(iii) For the number divisible by 5, the last digit in the humber should be 5. Remaining 4 places can be arranged from the remaining 4 digits accept digit 5.

∴ Total permutations = ¹P₁ × ⁴P₄

= 1 × 4!

= 1 × 24 = 24

In 5 letters of the word ROLLS, the letter L comes two times.

Total permutations = \(\frac{5!}{2!}=\frac{120}{2}\) = 60

In 5 letters of the word DOLLS, the letter L comes two times

Total permutations = \(\frac{5!}{2!}=\frac{120}{2}\) = 60

∴ The ratio of number of arrangements of all the letters of the word ROLLS and

DOLLS = 60 / 60 = 1 / 1 = 1 : 1

(2x + 3y)³

= ³C₀ (2x)³(3y)° + ³C₁ (2x)² (3y)¹ + ³C₂ (2x)(3y)² + ³C₃ (2x)⁰ (3y)³

= 8x³ + 3(4x²) (3y) + 3(2x) (9y²) + 27y³

= 8x³ + 36x²y + 54xy² + 27 y³

Given as

The word ‘STRANGE’

Here's 7 letters in the word ‘STRANGE’, which includes 2 vowels (A,E) and 5 consonants (S,T,R,N,G).

The vowels occupy only the odd places

Here are 7 letters in the word ‘STRANGE’. Out of these letters (A,E) are the vowels.

Here are 4 odd places in the word ‘STRANGE’. The two vowels can be arranged in ⁴P₂ ways.

The remaining 5 consonants an be arranged among themselves in ⁵P₅ ways.

Therefore, the total number of arrangements is

On using the formula,

P (n, r) = n!/(n – r)!

P (4, 2) × P (5, 5) = 4!/(4 – 2)! × 5!/(5 – 5)!

= 4!/2! × 5!

= (4 × 3 × 2!)/2! × 5!

= 4 × 3 × 5 × 4 × 3 × 2 × 1

= 12 × 120

= 1440

Thus, the number of arrangements therefore that the vowels occupy only odd positions is 1440.

Given : The word is ‘STRANGE.’ 

To find : A number of arrangements in which vowels come together 

Number of vowels in this word = 2(A, E) 

Now, 

Consider these two vowels as one entity

(AE together as a single letter) 

So, the total number of letters = 6(AE S T R N G) 

Formula used : 

Number of arrangements of n things taken all at a time = P(n, n)

P(n, r)  = \(\frac{n!}{(n-r)!}\)

∴ Total number of arrangements 

= the number of arrangements of 6 things taken all at a time 

= P(6, 6)

= \(\frac{6!}{(6-6)!}\)

= \(\frac{6!}{0!}\) 

{∵ 0! = 1} 

= 6! 

= 6 × 5 × 4 × 3 × 2 × 1 

= 720 

Two vowels which are together as a letter can be arranged in 2 

Ways like EA or AE 

Hence,

Total number of arrangements in which vowels come together = 2 × 720 = 1440

Permutation means an arrangement of things in which order of arrangement is important, while combination means selection of things in which order of selection is not important.

Option : (B)

We know, 

10 lacs = 10,00,000 and it has 7 places. 

The given numbers are 2, 3, 0, 3, 4, 2, 3. 

As the required number is greater than 10 lacs, so the 7^th place can be filled with all the digits except 0. 

So, the place can be filled in 6 ways. 

The other 6 places can be filled with other 6 digits, so the total number of ways to fill the remaining 6 places is = 6! 

But 2 is repeated twice and 3 is repeated thrice. 

So, total numbers greater than 10 lacs be formed from 2, 3, 0, 3, 4, 2, 3 is

= \(\frac{6\times 6!}{2!\times 3!}\) 

= \(\frac{6\times 720}{2!\times 6}\)

= 360.

We have to find the number of total outcomes when three dices are rolled the outcomes can be different and can be same.

We will use the concept of multiplication because there are three sub jobs dependent on each other and one job is performed one after the other. 

The choices of outcomes from one dice are six; there are three dices so choices for each dice will be six. 

The number of outcomes which are formed when three dices are rolled one after the other are 6 × 6 × 6 = 6³ = 216

We have to find the number of total outcomes when a coin is tossed three times, the outcomes can be different and can be the same. 

We will use the concept of multiplication because there are three sub jobs dependent on each other and one job is performed one after the other. 

The choices of outcomes from one coin when flipped once are two; the coin is flipped three times, so each time the number of outcomes will be two. 

The number of outcomes which are formed when a coin is flipped thrice, one after the other, are 

2 × 2 × 2 = 2³ = 8 

The number of outcomes which are formed when a coin is flipped four times, one after the other, are 

2 × 2 × 2 × 2 = 2⁴ = 16 

The number of outcomes which are formed when a coin is flipped five times, one after the other, are 

2 × 2 × 2 × 2 × 2 = 2⁵ = 32 

The number of outcomes which are formed when a coin is flipped n times, one after the other, are 

2 × 2 × 2………….till n times…… × 2 = 2ⁿ

There are 8 alphabets in the word EQUATION.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 8 different objects in 8 places is 

P(8,8) =  \(\frac{8!}{(8-8)!}\) = \(\frac{8!}{(0)!}\) = \(\frac{40320}{1}\) = 40320 

Hence there are 40320 words formed.

To find:

number of ways of hanging 6 pictures on 4 picture nails. There are 6 pictures to be placed in 4 places. 

Formula: 

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is 

P(n,r) = n!/(n-r)! 

Therefore, a permutation of 6 different objects in 4 places is 

P(6,4) = \(\frac{6!}{(6-4)!}\) =  \(\frac{6!}{2!}=\frac{720}{2}\) = 360 

This can be done by 360 ways