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Quantity A:

ORIENT

First we will fix the first LETTER: E, I & N then O will come;

No. of words possible with E as the first letter = 5! = 120

No. of words possible with I as the first letter = 5! = 120

No. of words possible with N as the first letter = 5! = 120

Now, we will fix first two letters: OE, OI and ON, then OR will come;

No. of words possible with OE as first two letters = 4! = 24

No. of words possible with OI as first two letters = 4! = 24

No. of words possible with ON as first two letters = 4! = 24

Now, we will fix first THREE letters: ORE, then ORI will come;

No. of words possible with ORE as first three letters = 3! = 6

Then the word ‘ORIENT’ will come (It will be the first word starting with ORI)

∴ RANK of ‘ORIENT’ in the dictionary = 360 + 72 + 6 + 1 = 439

Quantity B: 438

It is given that AMAN will not be a part of the main 5 if AKBAR is not in the main 5, but the vice versa does not apply, i.e. Akbar can still be a part of the main 5 even without Aman.

Hence, let us consider 3 cases:

Case 1: Both Aman and Akbar are in the main 5,

⇒ There are 10 more players left to fill 3 positions

⇒ No. of ways of selecting them = ¹⁰C₃ = 120

Case 2: Aman is not in the main 5, but Akbar is,

⇒ There are 10 more players left to fill 4 positions

⇒ No. of ways of selecting them = ¹⁰C₄ = 210

Case 3: NEITHER Aman nor Akbar is in the TEAM,

⇒ There are 10 more players left to fill 5 positions

⇒ No. of ways of selecting them = ¹⁰C₅ = 252

Quantity A:

One book for bookshelf 1 can be selected in ⁷C₁ ways.

Two books for bookshelf 2 can be selected in ⁶C₂ ways.

REMAINING books = 4 and Remaining bookshelves = 3.

Out of these 4 books, 1^st book can be PUT in any of these 3 bookshelves.

Similarly 2^nd book can be put in any of these 3 bookshelves.

3^rd book can be put in any of these 3 bookshelves.

4^th book can be put in any of these 3 bookshelves.

∴ These 4 books can be arranged in 3 × 3 × 3 × 3 = 81 ways

∴ Required number of ways

= ⁷C₁ × ⁶C₂ × 81

=7 × 15 × 81 = 8505

Quantity B:

Number of ways to choose 3 consonants and 2 vowels out of 5 consonants and 4 vowels

= ⁵C₃ × ⁴C₂ = 10 × 6 = 60

Number of ways in which 5 LETTERS can be arranged = 5! = 120

∴ Required number of words = 60 × 120 = 7200

The given WORD is TOUKIR. If vowels are arranged in dictionary order, then I will COME before O and O will come before U.

There are 6 letters in this word which need to be arranged.

Let’s first ARRANGE the vowels. They can be placed in any 3 of the 6 places.

Number of ways of choosing these places = ⁶C₃

There’s only one way of ARRANGING the vowels in the selected places, and that is the alphabetical order.

Now, the remaining three letters, i.e. T, K and R are to be placed in remaining 3 places. This can be done in 3! Ways.

∴ Total number of POSSIBLE arrangements = ⁶C₃ × 3! = 6 × 5 × 4 = 120

LET the vertices of the PENTAGON be A1, A2, A3, A4 and A5.

One triangle is formed by selecting a GROUP of 3 points from the 5 given vertices.

⇒ Number of triangles = Number of groups of 3 each from 5 points

⇒ ⁵C₃ = 5!/{(5 - 3)! × 3!} = 10

If three BALLS are PICKED, there can be balls of exactly two colours in the following cases:

i) Two blue balls and one yellow ball

This can be done in ⁷C₂ × ⁸C₁ ways, i.e. 21 × 8 = 168 ways.

ii) Two blue balls and one green ball

This can be done in ⁷C₂ × ⁵C₁ ways, i.e. 21 × 5 = 105 ways.

iii) Two yellow balls and one blue ball

This can be done in ⁸C₂ × ⁷C₁ ways, i.e. 28 × 7 = 196 ways.

iv) Two yellow balls and one green ball

This can be done in ⁸C₂ × ⁵C₁ ways, i.e. 28 × 5 = 140 ways.

v) Two green balls and one blue ball

This can be done in ⁵C₂ × ⁷C₁ ways, i.e. 10 × 7 = 70 ways.

vi) Two green balls and one yellow ball

This can be done in ⁵C₂ × ⁸C₁ ways, i.e. 10 × 8 = 80 ways.