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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.
The value of b6 is
Correct answer is '8'. Can you explain this answer?
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Most Upvoted Answer
Let an denote the number of all n-digit positive integers formed by t...
Solution:
Given, the digits used are 0, 1, or both.
Let an denote the number of n-digit positive integers formed by the digits 0, 1, or both such that no consecutive digits in them are 0.
Let bn be the number of such integers ending with digit 1 and cn be the number of such integers ending with digit 0.
For n = 1, a1 = 2, b1 = 1, c1 = 1.
For n = 2, a2 = 3, b2 = 1, c2 = 2.
For n > 2, the last digit can be either 0 or 1.
If the last digit is 1, then the second last digit can be either 0 or 1.
If the last digit is 0, then the second last digit can only be 1.
Thus, we get the following recursion relations:
an = bn + cn
bn = cn
cn = an-1
Using the above relations, we can compute a6 as follows:
a1 = 2
b1 = 1
c1 = 1
a2 = 3
b2 = 1
c2 = 2
a3 = 5
b3 = 2
c3 = 3
a4 = 8
b4 = 3
c4 = 5
a5 = 13
b5 = 5
c5 = 8
a6 = 21
b6 = 8
c6 = 13
Hence, the value of b6 is 8.
Given, the digits used are 0, 1, or both.
Let an denote the number of n-digit positive integers formed by the digits 0, 1, or both such that no consecutive digits in them are 0.
Let bn be the number of such integers ending with digit 1 and cn be the number of such integers ending with digit 0.
For n = 1, a1 = 2, b1 = 1, c1 = 1.
For n = 2, a2 = 3, b2 = 1, c2 = 2.
For n > 2, the last digit can be either 0 or 1.
If the last digit is 1, then the second last digit can be either 0 or 1.
If the last digit is 0, then the second last digit can only be 1.
Thus, we get the following recursion relations:
an = bn + cn
bn = cn
cn = an-1
Using the above relations, we can compute a6 as follows:
a1 = 2
b1 = 1
c1 = 1
a2 = 3
b2 = 1
c2 = 2
a3 = 5
b3 = 2
c3 = 3
a4 = 8
b4 = 3
c4 = 5
a5 = 13
b5 = 5
c5 = 8
a6 = 21
b6 = 8
c6 = 13
Hence, the value of b6 is 8.
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Community Answer
Let an denote the number of all n-digit positive integers formed by t...
To find b6, we have to find all 6 digit numbers ending with '1' such that no consecutive digits are '0'.
Some of the examples possible are:
1. 1 0 1 1 1 1
2. 1 0 1 0 1 1
3. 1 1 1 1 1 1
Three case possible:
1. One zero-It can be placed in any of the four places.
So, we get '4' such six digit numbers.
2. Two zeros-We get '3' such six digit numbers possible.
3. No zeros-We get only '1' such six digit number.
Hence, b6 = 4 + 3 + 1 = 8
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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.The value of b6 isCorrect answer is '8'. Can you explain this answer?
Question Description
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