/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Determine the number of \(n\)-di... [FREE SOLUTION] | 91Ó°ÊÓ

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Determine the number of \(n\)-digit quaternary \((0,1,2,3)\) sequences in which there is never a 3 anywhere to the right of a 0 .

Short Answer

Expert verified
To determine total number of \(n\)-digit quaternary sequences where there is never a 3 anywhere to the right of a 0, calculate number of valid sequences for each case and then add them up. Use the formulas \(a_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\), \(b_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\), \(c_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\), and \(d_{n}=a_{n-1}+b_{n-1}+d_{n-1}\) iterated from base cases, until the desired term is reached. Then add \(a_n + b_n + c_n + d_n\).

Step by step solution

01

Analyze the condition

Before starting, you should understand that having a 3 anywhere to the right of a 0 is not allowed. This condition means even 01, 02 and 03 can be formed, any combination ending with 0 can only be followed by 0,1 and 2 but never by 3. This restriction limits the possible quaternary numbers that can be produced.
02

Split into Cases

Given the restrictions, we can split our problem into four cases considering the end number of sequences by 0,1,2 and 3. If a sequence ends with 1,2, or 3, there are no limitations for previous numbers; however for the ones ending with 0, they can only be followed by sequences ending with 0,1 and 2 because of our condition which is not having 3 anywhere to the right of 0. Consequently, we will have 4 cases which can be described as sequences ending with 0,1,2,3.
03

Count the number of sequences for each case

Let's denote the number of n-digit sequences ending with 1,2,3, and 0 as \(a_n\), \(b_n\), \(c_n\), and \(d_n\) respectively. For a sequence to end in 1, 2, or 3, the previous sequence could be any \(n-1\) digit sequence, thus \(a_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\), \(b_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\), and \(c_{n}=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1}\). However, for a sequence to end in 0, the preceding sequence could not end in 3. Thus, \(d_{n}=a_{n-1}+b_{n-1}+d_{n-1}\).
04

Find the base cases

Observe that for the base case when \(n = 1\), \(a_1\), \(b_1\), \(c_1\), and \(d_1\) are all 1, since each of '1', '2', '3', and '0' are 1-digit quaternary number.
05

Iterate upon base cases to reach desired term

Applying the formulas derived from Step 3 and iterating from the base cases, you can get to the required term by applying the formulas repeatedly until \(n\) terms are reached.
06

Solution

Finally, add \(a_n\), \(b_n\), \(c_n\), and \(d_n\) to get the number of all admissible sequences. The final solution is \(a_n + b_n + c_n + d_n\).

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