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How many 4-digit positive integers exist that satisfy the following conditions: (A) Each of the first two digits must be 1, 4, or 5, and (B) the last two digits cannot be the same digit, and (C) each of the last two digits must be 5, 7, or 8?

 Sep 13, 2018
 #1
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0

3!  x   3! = 36 permutations.

 Sep 14, 2018
 #2
avatar+22182 
+9

How many 4-digit positive integers exist that satisfy the following conditions:

(A) Each of the first two digits must be 1, 4, or 5, and

(B) the last two digits cannot be the same digit, and

(C) each of the last two digits must be 5, 7, or 8?

 

(A)
Each of the first two digits must be 1, 4, or 5 ?

\(\begin{array}{|lllc|c|} \hline & x&x &xx \\ \hline &1 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & 3\times 100 \\\\ &4 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ &5 \cdots &\begin{Bmatrix} 1\\4\\5 \end{Bmatrix}\cdots &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} & +3\times 100 \\\\ \hline & & & & = 3\times 3\times 100 = {\color{red}900} \\ \hline \end{array} \)

 

There are 900 4-digit positive integers that satisfy the condition.

 

laugh

 Sep 14, 2018
edited by heureka  Sep 14, 2018
edited by heureka  Sep 14, 2018
 #3
avatar+22182 
+9

(B)
The last two digits cannot be the same digit ?

 

\(\begin{array}{|lcc|c|} \hline & xx &xx \\ \hline &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 00 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 11 & 90 \\\\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 22 & 90 \\\\ & \cdots \\ \\ &\begin{Bmatrix} 10\\ \vdots \\ 99 \end{Bmatrix}\cdots & 99 & 90 \\\\ \hline & & & = 90\times 10 = 900 \\ \hline \end{array}\)

 

1000 until 9999: There are 9000  four-digit integers.

 

9000 - 900 = 8100

 

There are 8100 4-digit positive integers that satisfy the condition.

 

laugh

 Sep 14, 2018
 #4
avatar+22182 
+8

(C)
Each of the last two digits must be 5, 7, or 8 ?

 

\(\begin{array}{|lclc|c|} \hline & x&x &xx \\ \hline &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 5& 100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 7& +100\times 3 \\\\ &\begin{Bmatrix} 00\\ \vdots \\ 99 \end{Bmatrix} \cdots &\begin{Bmatrix} 5\\7\\8 \end{Bmatrix}\cdots & 8& +100\times 3 \\\\ \hline & & & & = 3\times 100\times 3 = {\color{red}900} \\ \hline \end{array}\)

 

There are 900 4-digit positive integers that satisfy the condition.

 

laugh

 Sep 14, 2018

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