(2)

No. of words formed when word starts from B = 5!

No. of words formed when word starts from C = 5!

No. of words formed when word starts from I = 5!

No. of words formed when word starts from L = 5!

No. of words formed when word starts from PB = 4!

No. of words formed when word starts from PC = 4!

No. of words formed when word starts from PI = 4!

No. of words formed when word starts from PL = 4!

No. of words formed when word starts from PUBC = 2!

No. of words formed when word starts from PUBI = 2!

No. of words formed when word starts from PUBLC = 1

Now, the word comes is PUBLIC = 1

∴ Serial number of word PUBLIC

= 4 × 5! + 4 × 4! + 2 × 2! + 1 + 1 = 582

(2)

(2)

Number of girls = 5

Number of boys = 7

The number of ways of arranging boys around a table is (7 – 1)! = 6!

Now, there are 7 spaces in between two boys, so 5 girls arranged in 7 gaps by ${}^7{P}_5$ ways.

So, required number of ways = $6!\times {}^7{P}_5=126\times {(5!)}^2$

(1)

In the word ''MATHEMATICS'' there are, $2\to \text{M's; }2\to \text{A's; }2\to \text{T's}$

Total number of words formed by the letters of the word ''MATHEMATICS'' is $\frac{\lfloor 11}{\lfloor 2\times \lfloor 2\times \lfloor 2}$

Number of words in which 'C' and 'S' come together is $\frac{\lfloor 10}{\lfloor 2\times \lfloor 2\times \lfloor 2}\times 2$

Then, number of words in which 'C' and 'S' do not come together is $\frac{\lfloor 11}{\lfloor 2\times \lfloor 2\times \lfloor 2}-\frac{2\lfloor 10}{\lfloor 2\times \lfloor 2\times \lfloor 2}=\frac{9\times \lfloor 10}{2\times 2\times 2}$

$\Rightarrow \frac{9\times 10\times 9\times 8\times 7\times 6!}{2\times 2\times 2}=6!\times k (\because \text{Given})\Rightarrow k=5670$

(3)

8 persons can be divided into 3 groups of 3, 3, and 2 members each. 

$\therefore$  $\text{Required number of ways}=\frac{8!}{3! 3! 2! 3!}\times 3!$

$=\frac{8\times 7\times 6\times 5\times 4}{4}=56\times 30=1680$

(1)

Given word is MATHS.

In alphabetical order: A, H, M, S, T

Now, word starting with

A_ _ _ _ i.e., 4! ; H_ _ _ _  i.e., 4!

M_ _ _ _ i.e., 4! ; S_ _ _ _  i.e., 4!

T A_ _ _  i.e., 3! ; T H A M S i.e., 1

$\therefore$   $\text{Rank of the word THAMS}=4\times 4!+1\times 3!+1=103$

(2)

Numbers greater than 40000 and divisible by 5 can be formed using digits 0, 1, 3, 5, 7, 9 in the following ways:

Total number of ways = 24 + 48 + 48 = 120

(1)

Given, A, D, M, N, O, Y

A → 5! ; D → 5! ; M A → 4! ; M N → 4!

M O A → 3! ; M O D → 3! ; M O N A → 2! ; M O N D A Y

⇒ Rank = 2(5)! + 3(4)! + 2(3)! + 2! + 1

             = 240 + 72 + 12 + 2 + 1 = 327

(3)

Case I: All the three digits are same i.e., 111, 333, 555, 888. All the numbers are divisible by 3.

Case II: Two digits are same.

We can see that when two digits are same, then numbers having digits 5, 5, 8 or 8, 8, 5 will be divisible by 3 as their sum is divisible by 3.

Now, for each number, we have $\frac{3!}{2!}=3$ arrangements.

$\therefore$    In this case, we have 6 numbers divisible by 3.

Case III: All digits are distinct.

i.e., (1, 3, 5), (1, 3, 8)

So, number formed = 2 × 3! = 12

Hence, total 3-digit numbers formed by digits 1, 3, 5, 8 and divisible by 3 is 4 + 6 + 12 = 22.

(2)

In first column, 1 can be placed in any of the 5 places = 5

In second column, 1 can be placed in any of the 4 places = 4

In third column, 1 can be placed in any of the 3 places = 3

In fourth column, 1 can be placed in any of the 2 places = 2

In fifth column, 1 can be placed in any of the 1 place = 1

Required number of ways = 5 × 4 × 3 × 2 × 1 = 120