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Let an 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 bn = the number of such n-digit integers ending with digit 1 and cn = the number of such n-digit integers ending with digit 0. [JEE 2012]
The value of b6 is
- a)7
- b)8
- c)9
- d)11
Correct answer is option 'B'. Can you explain this answer?
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Solution:
Let us first find a recursive formula for a_n, 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.
Consider an (n+1)-digit integer formed by appending a digit to an n-digit integer. If the last digit of the n-digit integer is 1, then we can append either 0 or 1 to it. If the last digit of the n-digit integer is 0, then we can only append 1 to it. If the last digit of the n-digit integer is neither 0 nor 1, then we can append either 0 or 1 to it. Thus, we have
a_n+1 = a_n + b_n + c_n
where b_n is the number of n-digit integers ending with digit 1 and c_n is the number of n-digit integers ending with digit 0.
Now, let us find a recursive formula for b_n and c_n. Consider an (n+1)-digit integer ending with digit 1. If the second-to-last digit is 1, then we can append any of the three digits 0, 1, or 1 to it. If the second-to-last digit is 0, then we can only append 1 to it. If the second-to-last digit is neither 0 nor 1, then we can append any of the two digits 0 or 1 to it. Thus, we have
b_n+1 = a_n + b_n
Similarly, for c_n+1, we have
c_n+1 = a_n + c_n
Using these recursive formulas, we can compute b_6 and c_6 as follows:
a_1 = 2 (the two digits 0 and 1 are allowed)
b_1 = 1 (only the digit 1 is allowed)
c_1 = 1 (only the digit 0 is allowed)
a_2 = 3 (the three digits 00, 01, and 10 are allowed)
b_2 = 1 (only the digit 1 can follow 1)
c_2 = 1 (only the digit 0 can follow 10)
a_3 = 5 (the five digits 000, 001, 010, 100, and 101 are allowed)
b_3 = 2 (the digits 11 and 101 are allowed)
c_3 = 2 (the digits 00 and 100 are allowed)
a_4 = 9
b_4 = 4
c_4 = 4
a_5 = 16
b_5 = 7
c_5 = 9
a_6 = 28
b_6 = 8
c_6 = 13
Therefore, the number of 6-digit integers ending with digit 1 is b_6 = 8. The correct answer is option B.
Let us first find a recursive formula for a_n, 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.
Consider an (n+1)-digit integer formed by appending a digit to an n-digit integer. If the last digit of the n-digit integer is 1, then we can append either 0 or 1 to it. If the last digit of the n-digit integer is 0, then we can only append 1 to it. If the last digit of the n-digit integer is neither 0 nor 1, then we can append either 0 or 1 to it. Thus, we have
a_n+1 = a_n + b_n + c_n
where b_n is the number of n-digit integers ending with digit 1 and c_n is the number of n-digit integers ending with digit 0.
Now, let us find a recursive formula for b_n and c_n. Consider an (n+1)-digit integer ending with digit 1. If the second-to-last digit is 1, then we can append any of the three digits 0, 1, or 1 to it. If the second-to-last digit is 0, then we can only append 1 to it. If the second-to-last digit is neither 0 nor 1, then we can append any of the two digits 0 or 1 to it. Thus, we have
b_n+1 = a_n + b_n
Similarly, for c_n+1, we have
c_n+1 = a_n + c_n
Using these recursive formulas, we can compute b_6 and c_6 as follows:
a_1 = 2 (the two digits 0 and 1 are allowed)
b_1 = 1 (only the digit 1 is allowed)
c_1 = 1 (only the digit 0 is allowed)
a_2 = 3 (the three digits 00, 01, and 10 are allowed)
b_2 = 1 (only the digit 1 can follow 1)
c_2 = 1 (only the digit 0 can follow 10)
a_3 = 5 (the five digits 000, 001, 010, 100, and 101 are allowed)
b_3 = 2 (the digits 11 and 101 are allowed)
c_3 = 2 (the digits 00 and 100 are allowed)
a_4 = 9
b_4 = 4
c_4 = 4
a_5 = 16
b_5 = 7
c_5 = 9
a_6 = 28
b_6 = 8
c_6 = 13
Therefore, the number of 6-digit integers ending with digit 1 is b_6 = 8. The correct answer is option B.
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Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer?
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Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer?.
Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let andenote 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 bn= the number of such n-digit integers ending with digit 1 and cn= the number of such n-digit integers ending with digit 0. [JEE 2012]The value of b6isa)7b)8c)9d)11Correct answer is option 'B'. Can you explain this answer?.
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