Let a_n denote the number of all...

MathematicsFundamental Principle of CountingJEE Advanced 2012Moderate

Comprehension Passage

Let

a_{n}

denote the number of all

n

-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let

b_{n}

= the number of such

n

-digit integers ending with digit 1 and

c_{n}

= the number of such

n

-digit integers ending with digit 0.

Question 1:

The value of $b_{6}$ is

(A)

(B)

(C)

(D)

Answer:B

Question 2:

Which of the following is correct?

(A)

a_{17} = a_{16} + a_{15}

(B)

c_{17} \neq = c_{16} + c_{15}

(C)

b_{17} \neq = b_{16} + c_{16}

(D)

a_{17} = c_{17} + b_{16}

Answer:A

Visualized Solution (Hindi)

Understanding the Constraints

Digits allowed: $0, 1$
Condition: No consecutive $0$ s (No ' $00$ ')
Constraint: First digit must be $1$ (for an $n$ -digit positive integer)

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{n} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c_{n}

$a_{n}$ : Total number of such $n$ -digit integers
$b_{n}$ : Number of such integers ending with $1$
$c_{n}$ : Number of such integers ending with $0$
Total relation: $a_{n} = b_{n} + c_{n}$

Deriving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{n} = a_{n - 1}

Case $b_{n}$ : Ends in $1$
The first $n - 1$ digits must form a valid $(n - 1)$ -digit integer.
Therefore, $b_{n} = a_{n - 1}$

Deriving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c_{n} = a_{n - 2}

Case $c_{n}$ : Ends in $0$
The $(n - 1)$ -th digit MUST be $1$ to avoid ' $00$ '.
The first $n - 2$ digits must form a valid $(n - 2)$ -digit integer.
Therefore, $c_{n} = b_{n - 1} = a_{n - 2}$

The Recurrence Relation

Substitute $b_{n}$ and $c_{n}$ into the total sum:
a_n = a_{n-1} + a_{n-2}
This holds for $n \geq 3$ .

Base Cases: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{2}

For $n = 1$ : {1} $⟹ a_{1} = 1$
For $n = 2$ : {11, 10} $⟹ a_{2} = 2$

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{3} to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{6}

$a_{3} = a_{2} + a_{1} = 2 + 1 = 3$
$a_{4} = a_{3} + a_{2} = 3 + 2 = 5$
$a_{5} = a_{4} + a_{3} = 5 + 3 = 8$
$a_{6} = a_{5} + a_{4} = 8 + 5 = 13$

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{6}

We know $b_{n} = a_{n - 1}$
Substitute $n = 6$ : $b_{6} = a_{5}$
From our previous step, $a_{5} = 8$
$∴ b_{6} = 8$

Verifying the Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{17}

The general recurrence is $a_{n} = a_{n - 1} + a_{n - 2}$
For $n = 17$ : $a_{17} = a_{16} + a_{15}$
This matches the first option.
Key Takeaway: The sequence $a_{n}$ follows the Fibonacci recurrence starting from $a_{1} = 1, a_{2} = 2$ .

00:00 / 00:00

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Comprehension Passage

The value of $b_{6}$ is

Which of the following is correct?

Visualized Solution (Hindi)

Understanding the Constraints

Defining $b_{n}$ and $c_{n}$

Deriving $b_{n} = a_{n - 1}$

Deriving $c_{n} = a_{n - 2}$

The Recurrence Relation

Base Cases: $a_{1}$ and $a_{2}$

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of b6​ is

Which of the following is correct?

Visualized Solution (Hindi)

Understanding the Constraints

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cn​

Deriving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​=an−1​

Deriving .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cn​=an−2​

The Recurrence Relation

Base Cases: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a3​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a6​

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b6​

Verifying the Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a17​

Conceptually Similar Problems

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=43​−(43​)2+(43​)3+⋯+(−1)n−1(43​)n and bn​=1−an​, then find the least natural number n0​ such that bn​>an​ for all n≥n0​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S1​=∑j=110​j(j−1)10Cj​,S2​=∑j=110​j10Cj​ and S_3 = \sum_{j=1}^{10} j^2 ^{10}C_j. Statement-1: S3​=55×29. Statement-2: S1​=90×28 and S2​=10×28.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, let a(n)=1+21​+31​+41​+⋯+(2n)−11​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integers n1​,n2​, the value of the expression (1+i)n1​+(1+i3)n1​+(1+i5)n2​+(1+i7)n2​, where i=−1​ is a real number if and only if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of (030​)(1030​)−(130​)(1130​)+(230​)(1230​)−....+(2030​)(3030​) is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of b6​ is

Which of the following is correct?

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If an​=43​−(43​)2+(43​)3+⋯+(−1)n−1(43​)n and bn​=1−an​, then find the least natural number n0​ such that bn​>an​ for all n≥n0​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S1​=∑j=110​j(j−1)10Cj​,S2​=∑j=110​j10Cj​ and S_3 = \sum_{j=1}^{10} j^2 ^{10}C_j. Statement-1: S3​=55×29. Statement-2: S1​=90×28 and S2​=10×28.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For a positive integer n, let a(n)=1+21​+31​+41​+⋯+(2n)−11​. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For positive integers n1​,n2​, the value of the expression (1+i)n1​+(1+i3)n1​+(1+i5)n2​+(1+i7)n2​, where i=−1​ is a real number if and only if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of (030​)(1030​)−(130​)(1130​)+(230​)(1230​)−....+(2030​)(3030​) is

The value of $b_{6}$ is

Defining $b_{n}$ and $c_{n}$

Deriving $b_{n} = a_{n - 1}$

Deriving $c_{n} = a_{n - 2}$

Base Cases: $a_{1}$ and $a_{2}$

Calculating $a_{3}$ to $a_{6}$

Finding $b_{6}$

Verifying the Relation for $a_{17}$

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j^{10} C_{j}$ and $S_3 = \sum_{j=1}^{10} j^2 ^{10}C_j$ . Statement-1: $S_{3} = 55 \times 2^{9}$ . Statement-2: $S_{1} = 90 \times 2^{8}$ and $S_{2} = 10 \times 2^{8}$ .

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

For positive integers $n_{1}, n_{2}$ , the value of the expression $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$ , where $i = - 1$ is a real number if and only if

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is

The value of $b_{6}$ is

If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Let $S_{1} = \sum_{j = 1}^{10} j (j - 1)^{10} C_{j}, S_{2} = \sum_{j = 1}^{10} j^{10} C_{j}$ and $S_3 = \sum_{j=1}^{10} j^2 ^{10}C_j$ . Statement-1: $S_{3} = 55 \times 2^{9}$ . Statement-2: $S_{1} = 90 \times 2^{8}$ and $S_{2} = 10 \times 2^{8}$ .

For a positive integer $n$ , let $a (n) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{( 2 ^{n} ) - 1}$ . Then

For positive integers $n_{1}, n_{2}$ , the value of the expression $(1 + i)^{n_{1}} + (1 + i^{3})^{n_{1}} + (1 + i^{5})^{n_{2}} + (1 + i^{7})^{n_{2}}$ , where $i = - 1$ is a real number if and only if

The value of $(0 30) (10 30) - (1 30) (11 30) + (2 30) (12 30) - .... + (20 30) (30 30)$ is