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"SUCCESS", there are 3 S, 2 C , 1 U & 1 E.

So, there are 7!/(3! 2!) = (7*6*5*4)/2 = 420 possible arrangements

When all 3 S come together , there are (1+2+1+1)!/2! = 5!/2! = 60

Number of maths book = 4 

Number of physics books = 3 

Number of chemistry books = 2 

Number of biology books = 1 

Since we want books of the same subjects together, we have to treat all maths books as 1 unit, all physics books as 1 unit, all chemistry books as 1 unit and all biology books as 1 unit. Now total number of units = 4 

They can be arranged in 4! ways. After this arrangement. 

4 maths book can be arranged in 4! ways 

3 physics book can be arranged in 3! ways 

2 chemistry book can be arranged in 2! ways and 1 biology book can be arranged in 1! way 

∴ Total Number of arrangements 4! 4! 3! 2! = 6912

Selecting a chair person from the 10 persons can be done in 10 ways 

After the selection of chair person only 9 persons are left out so selecting a secretary (from the remaining a persons) can be done in 9 ways.

The remaining persons = 8 

Totally we need to select 6 persons 

We have selected 2 persons. 

So we have to select 4 persons 

Selecting 4 from 8 can be done in ⁸C₄ ways

∴ Total number of selection = 10 x 9 x ⁸C₄

= (10 x 9 x 8 x 7 x 6 x 5)/(4 x 3 x 2 x 1) = 6300

(b) P_{n + 1} – 1 

1 + 1! + 2! + 3! + … + n! 

Now 1 + 1 (1!) = 2 = (1 + 1)! 

1 + 1 (1!) + 2(2!) = 1 + 1 + 4 = 6 = 3! 

1 + 1(1!) + 2(2!)+ 3(3!) = 1 + 1 + 4 + 18 = 24 = 4! 

1 + 1(1!) + 2(2!) + 3(3!) ….+ n(n!) = (n + 1)! – 1 

= ^{n + 1}P_{n + 1} – 1 = P_{n + 1} – 1

(b) 3

a₂ – a = 2 + 4 = 6 

a₂ – a – 6 = 0 

(a – 3) (a + 2) = 0 ⇒ a = 3

P(n) is the statement 5^{n + 1} + 4 × 6ⁿ – 9 is ÷ by 20 

P(1) = 5^{1 + 1} + 4 × 6¹ – 9 = 5² + 24 – 9 

= 25 + 24 – 9 = 40 ÷ by 20 

So P(1) is true 

Assume that the given statement is true for n = k 

(i.e) 5^{k + 1} + 4 × 6ⁿ – 9 is ÷ by 20 

P(1) = 5^{1 - 1} + 4 × 6¹ – 9 

= 25 + 24 – 9 

So P(1) is true 

To prove P(k + 1) is true 

P(k + 1) = 5^{k + 1 + 1} + 4 × 6^{k + 1 + 1} – 9 

= 5 × 5^{k + 1} + 4 × 6 × 6^k – 9 

= 5[20C + 9 – 4 × 6^k] + 24 × 6^k – 9 [from(1)] 

= 100C + 45 – 206^k + 246^k – 9 

= 100C + 46^k + 36

= 100C + 4(9 + 6^k) 

Now for k = 1 ⇒ 4(9 + 6^k) = 4(9 + 6) 

= 4 × 15 = 60 ÷ by 20 . 

For k = 2 = 4(9 + 6²) = 4 × 45 = 180 ÷ 20 

So by the principle of mathematical induction 4(9 + 6^k) is ÷ by 20 

Now 100C is ÷ by 20. 

So 100C + 4(9 + 6^k) is ÷ by 20 

⇒ P(k + 1) is true whenever P(k) is true. So by the principle of mathematical induction P(n) is true.

P(n) is the statement 10ⁿ + 3 × 4^{n + 2} + 5 is ÷ by 9 

P(1) = 10¹ + 3 × 4² + 5 = 10 + 3 × 16 + 5 

= 10 + 48 + 5 = 63 ÷ by 9

So P(1) is true. Assume that P(k) is true 

(i.e.) 10^k + 3 × 4^{k + 2} + 5 is ÷ by 9 

(i.e.) 10^k + 3 × 4^{k + 2} + 5 = 9C (where C is an integer) 

⇒ 10^k = 9C – 5 – 3 × 4^{k + 2} ….. (1) 

To prove P(k + 1) is true. 

Now P(k + 1) = 10^{k + 1} + 3 × 4^{k + 3} + 5 

= 10 × 10^k + 3 × 4 × 4^{k + 2} + 5 

= 10[9C – 5 – 3 × 4^{k + 2}] + 3 × 4^{k + 2} × 4 + 5 

= 10[9C – 5 – 3 × 4^{k + 2}] + 12 × 4^{k + 2} + 5 

= 90C – 50 – 30 × 4^{k + 2} + 12 × 4^{k + 2} + 5 

= 90C – 45 – 18 × 4^{k + 2} 

= 9[10C – 5 – 2 × 4^{k + 2}] which is ÷ by 9 

So P(k + 1) is true whenever P(K) is true. So by the principle of mathematical induction P(n) is true.

(a) r!

1(2) (3) … (r) = r! which is ÷ by r!

Let P(n) : n³ – n

Step 1 : 

P(2): 2³ – 2 = 6 which is divisible by 6. So it is true for P(2). 

Step 2 : 

P(A): k³ – k = 6λ. Let it is be true for k ≥ 2 

⇒ k³ = 6λ + k … (i)

Step 3 : 

P(k + 1) = (k + 1)³ – (k + 1) 

= k³ + 1 + 3k² + 3k – k – 1 = k³ – k + 3(k² + k) 

= k³ – k + 3(k² + k) = 6λ + k – k + 3(k² + k) 

= 6λ + 3(k² + k) [from (i)] 

We know that 3(k² + k) is divisible by 6 for every value of k ∈ N. 

Hence P(k + 1) is true whenever P(k) is true.

(i) To find the number of words starting with L 

Number of letters in LOTUS = 5 when the first letter is L it can be filled in 1 way only. So the remaining 4 letters can be arranged in 4! =24 ways = n(A). 

When the last letter is S it can be filled in the 1 way and the remaining 4 letters can be arranged is 4! = 24 ways = n(B)

Now the number of words starting with L and ending with S is L OTU S

(1) (1) 3! = 6 = n(A ∩ B) 

Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 

= 24 + 24 – 6 = 42 

Now, neither words starts with L nor ends with S = 42

(ii) Number of letters of the word LOTUS = 5. 

They can be arranged in 5 ! = 120 ways

Number of words starting with L and ending with S = 42 

So the number of words neither starts with L nor ends with S = 120 – 42 = 78

(i) The given digits are 0, 1, 2, 3, 4, 5 

We have to find numbers between 100 and 500. So the 100’s place can be filled (by the numbers 1, 2, 3, 4) in 4 ways. 

The 10’s place can be filled in (using 0, 1, 2, 3, 4, 5) 6 ways and the unit-place can be filled in (using 0,1, 2, 3, 4, 5) 6 ways 

But the number 100 should be excluded 

So the number of numbers between 100 and 500 = 4 × 6 × 6 = 144

(ii) The 100’s place can be filled (by using 1, 2, 3, 4) 10’s in 4 ways 

The 10’s place can be filled in (6 – 1) 5 ways and the unit place can be filled in (5 – 1) 4 ways 

So the number of 3 digit number 4 × 5 × 4 = 80

From 1 to 1000, the numbers ÷ by 2 = 500 the number ÷ by 5 = 200 and the numbers ÷ by 10 = 100(5 × 2 = 10) 

So number ÷ by 2 or 5 = 500 + 200 – 100 = 600 

Total numbers from 1 to 1000 = 1000 

So the number of numbers which are ÷ neither by 2 nor by 5 = 1000 – 600 = 400 

(i) Number of digit given = 4 (2,4, 6, 8) 

So the unit place can be filled in 4 ways, 10’s place can be filled in 4 ways and 100’s place can be filled in 4 ways 

∴ The unit place, 10’s place and 100’s place together can be filled (i.e) So the Number of 3 digit numbers = 4 × 4 × 4 = 64 ways

(ii) The unit place can be filled (using the 4 digits) in 4 ways after filling the unit place since repetition of digits is not allowed that digit should be excluded. So the 10’s place can be filled in (4 – 1) 3 ways and the 100’s place can be filled in (3 – 1) 2 ways 

So the unit place, 10’s and 100’s places together can be filled in 4 × 3 × 2 = 24 ways 

(i.e) The number of 3 digit numbers = 4 × 3 × 2 = 24 ways

(i) The unit place is filled (by 3) in 1 way 

The 10’s place can be filled in 10 ways 

The 100’s place can be filled in 9 ways (excluding 0) 

So the number of 3 digit numbers with 3 unit – place = 9 × 10 × 1 = 90

(ii) The digits are 0, 1,2, 3, 4, 5, 6, 7, 8, 9 

A three digit number has 3 digits l’s, 10’s and 100’s place. 

The unit place is (filled by 3) filled in one way. 

After filling the unit place since the digit ‘0’ is there, we have to fill the 100’s place. Now to fill the 100’s place we have 8 digits (excluding 0 and 3) 

So 100’s place can be filled in 8 ways. 

Now to fill the 10’s place we have again 8 digits (excluding 3 and any one of the number) 

So 10’s place can be filled in 8 ways. 

∴ Number of 3 digit numbers with ‘3’ in unit place 

= 8 × 8 × 1 = 64

(i) The given digits are 0, 1, 2, 3, 4, 5. A number will be divisible by 5 if the digit in the unit place is 0 or 5 

So the unit place can be filled by 0 or 5

(a) When the unit place is 0 it is filled in 1 way 

And so 10’s place can be filled in 5 ways (by using 1, 2, 3, 4, 5) and 100’s place can be filled in (5 – 1) 4 ways 

So the number of 3 digit numbers with unit place 0 

= 1 × 5 × 4 = 20

(b) When the unit place is 5 it is filled in 1 way 

Since 0 is given as a digit to fill 100’s place 0 should be excluded 

So 100’s place can be filled in (excluding 0 and 5) 4 ways and 10’s place can be filled in (excluding 5 and one digit and including 0) 4 ways 

So the number of 3 digit numbers with unit place 5 = 1 × 4 × 4 = 16 

∴ Number of 3 digit numbers ÷ by 5 = 20 + 16 = 36

(ii) The digits are 0 1 2 3 4 5 

To get a number divisible by 5 we should have the unit place as 5 or 0 

So the unit place (using 0 or 5) can be filled in 2 ways. 

The 10’s place can be filled (Using 0, 1, 2, 3, 4, 5) in 6 ways and the 100’s place (Using 1, 2, 3, 4, 5) can be filled in 5 ways. 

So the number of 3 digit numbers ÷ by 5 (with repetition) = 2 × 6 × 5 = 60

(i) We have to find 4 digit numbers 

The 1000’s place can be filled in 9 ways (excluding zero) and the 100’s, 10’s and unit places respectively can be filled in 10, 10, 10 ways (including zero) 

So the number of numbers between 999 and 10000 = 9 × 10 × 10 × 10 = 9000

(ii) Since 0 is given as a digit we have to start filling 1000’s place. 

Now 1000’s place can be filled in 9 ways (excluding 0) 

Then the 100’s place can be filled in 9 ways (excluding one digit and including 0) 

10’s place can be filled in (9 – 1) 8 ways and unit place can be filled in (8 – 1) 7 ways So the number of 4 digit numbers are 9 × 9 × 8 × 7 = 4536 ways

(iii) Required number of numbers = 9000 – 4536 = 4464 numbers

(c) ¹²C₈ – ¹⁰C₆

Number of way of selecting 8 people from 12 in ¹²C₈ 

∴ out of remaining people 8 can attend in ¹⁰C₈ 

The number of ways in which two of them do not attend together = ¹²C₈ – ¹⁰C₆

There are 4 odd places in the word

∴ Number of permutations = ⁴P₂ = 4!/2! = 12

Now the remaining 6 places filled with remaining 6 letters 

∴ Number of permutations = ⁶P₆ = 6!/0! = 720

Hence the total number of permutations = 12 × 720 = 8640

SIMPLE 

Total Number of letters = 6 

They can be arranged in 6! ways

∴ Number of words = 6!

= 6 × 5 × 4 × 3 × 2 × 1 = 720

No. of cards = 52 : In that number of aces = 4 

No. of cards needed = 5

In that 5 cards number of aces needed = 3 

So the 3 aces can be selected from 4 aces in ⁴C₃ = ³C₁ = 4 ways 

So the remaining = 5 – 3 = 2 

This 2 cards can be selected in ⁴⁸C₂ ways

⁴⁸C₂ = (48 x 47)/(2 x 1) = 1128 ways

∴ No. of ways in which the 5 cards can be selected = (⁴⁸C₂)(⁴C₃) 

= (48 x 47)/(2 x 1) = 4 = 4512

P(n) : (1 + xⁿ) ≥ 1 + nx 

P(1) : (1 + x)¹ ≥ 1 + x 

⇒ 1 + x ≥ 1 + x, which is true. 

Hence, P(1) is true. 

Let P(k) be true 

(i.e.) (1 + x)^k ≥ 1 + kx 

We have to prove that P(k + 1) is true. 

(i.e.) (1 + x)^{k + 1} ≥ 1 + (k + 1)x 

Now, (1 + x)^{k + 1} ≥ 1 + kx [∵ p(k) is true] 

Multiplying both sides by (1 + x), we get 

(1 + x)^k (1 + x) ≥ (1 + kx)(1 + x) 

⇒ (1 + x)^{k + 1} ≥ 1 + kx + x + kx²

⇒ (1 + x)^{k + 1} ≥ 1 + (k + 1)x + kx² ... (1) 

Now, 1 + (k + 1) x + kx₂ ≥ 1 + (k + 1)x … (2) [∵ kx > 0] 

From (1) and (2), we get 

(1 + x)^{k + 1} ≥ 1 + (k + 1)x 

∴ P(k + 1) is true if P(k) is true. 

Hence, by the principle of mathematical induction,

P(n) is true for all values, of n.

Here total number of digits = 6 

The unit place can be filled with any one of the digits 2, 4, 6.

So number of permutation = ³P₁ = 3!/2! = 3

Now the tens and hundred place can be filled by remaining 5 digits.

So number of permutation = ⁵P₃ = 5!/3! = (5 x 4 x 3!)/3! = 20

Hence total number of permutations = 3 x 20 = 60

Any number divisible by 5, its unit place must have 0 or 5. We have to find 4-digit number greater than 6000 and less than 7000. 

So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed. 

∴ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.

But the required number is greater than 6000 and less than 7000. So, thousand place can be filled with 1 digits (i.e) 6

So, the total number of integers = 1 × 8 × 7 × 2 = 112 

Hence, the required number of integers = 112

Given that all the 5 digit numbers are greater than 7000. 

So, the ways of forming 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120 

Now, all the four digit number greater than 7000 can be formed as follows. 

Thousand place can be filled with 3 ways 

Hundred place can be filled with 4 ways 

Tenths place can be filled with 3 ways 

Units place can be filled with 2 ways 

So, the total number of 4-digits numbers = 3 × 4 × 3 × 2 = 72 

∴ Total number of integers = 120 + 72 = 192 

Hence, the required number of integers = 192