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Bekah has four brass house number digits: 2, 3, 5 and 7, and only has one of each number. How many distinct numbers can she form using one or more of the digits?

 Nov 5, 2018
 #1
avatar+98172 
+2

Using one digit, we have 4 possible numbers

 

Using two of the digits, we have  P (4,2)  = 12 possible numbers

 

Using therr of the digits, we have P(4,3)  = 24 possible numbers

 

Using all four, we have 4!  = 24 possible numbers

 

So....the total number of possibilities is

 

4 + 12 + 24 + 24   =

 

64 possibilities

 

 

 

cool cool cool

 Nov 5, 2018
 #2
avatar+21848 
+10

Bekah has four brass house number digits:
2, 3, 5 and 7, and only has one of each number.
How many distinct numbers can she form using one or more of the digits?

 

\(\text{binary code} \\ \begin{array}{|c|c|c|c|r|c|} \hline 2 & 3 & 5 & 7 & & \text{distinct numbers} \\ \hline 0&0&0&1 & 7 & 1! \\ 0&0&1&0 & 5 & 1! \\ 0&0&1&1 & 57 \text{ or } 75 & 2! \\ 0&1&0&0 & 3 & 1! \\ 0&1&0&1 & 37 \text{ or } 73 & 2! \\ 0&1&1&0 & 35 \text{ or } 53 & 2! \\ 0&1&1&1 & 357 & 3! \\ 1&0&0&0 & 2 & 1! \\ 1&0&0&1 & 27 \text{ or } 72 & 2! \\ 1&0&1&0 & 25 \text{ or } 52 & 2! \\ 1&0&1&1 & 257 & 3! \\ 1&1&0&0 & 23 \text{ or } 32 & 2! \\ 1&1&0&1 & 237 & 3! \\ 1&1&1&0 & 235 & 3! \\ 1&1&1&1 & 2357 & 4! \\ \hline &&&&\text{sum}& = 4\times 1! + 6\times 2! + 4\times 3!+1\times 4! \\ &&&&& = \dbinom{4}{1}1! +\dbinom{4}{2}2! +\dbinom{4}{3}3! +\dbinom{4}{4}4! \\ &&&&& = 1!C_4^1 +2!C_4^2 + 3!C_4^3 + 4!C_4^4 \\ &&&&& = 4+12+24+24 \\ &&&&&\mathbf{ = 64} \\ \hline \end{array}\)

 

laugh

 Nov 6, 2018
edited by heureka  Nov 8, 2018

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