<span><span>← BACKPRINT<span>+ TEXT SIZE –</span>SEARCHHOME</span><span>Question from Mike, a parent:Sorry if you have answered this before but I don't understand the explanations I have seen.
How many possible combinations are in a 5 digit code (like a Zip Code). Is there an equation that can be used for this type of question?All the best and thank you,
mike</span>Hi </span><span>I am going to assume there are no restrictions on the 5 digits, in particular 00070 and 00000 are allowed.First let's look atHow many possible combinations are in a 1 digit codeThe answer is 10. They are 0, 1, 2, 3, 4, 5, 6 ,7 ,8 and 9Now what aboutHow many possible combinations are in a 2 digit codeEach of the 1 digit codes can be transformed into a 2 digit code by appending a second digit. For example 0 can be transformed into a 2 digit code in 10 ways. they are 00, 01, 02, 03, 04, 05, 06, 07, 08 and 09. Likewise 1 can be transformed into a 2 digit code in 10 ways. Similarly for all ten 1 digit codes and hence there are 10 × 10 = 102 = 100 possible 2 digit codes.How many possible combinations are in a 3 digit codeEach of the one-hundred 2 digit codes can be transformed into a 3 digit code by appending a third digit. I hope it is clear now that there are 10 × 10 × 10 = 103 = 1000 possible 3 digit codes. Likewise there are 104 possible 4 digit codes are <span>105 = 100,000</span> possible 5 digit codes.In Canada where I live our postal codes have a different structure. They are letter, digit, letter (space) digit, letter, digit. For example the postal code of post office in the town of Fort Qu'Appelle Saskatchewan is S0G 1S0. With this structure there are 26 choices for the first character, 10 choices for the second character, 26 choices for the third character and so on. In total there are 26 × 10 × 26 × 10 × 26 × 10 = 17,576,000 possible Canadian postal codes if there are no restrictions on what letters and digits can be used.I hope this helps,
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Answer:
= 90 - 3n
Step-by-step explanation:
There is a common difference between consecutive terms, that is
84 - 87 = 81 - 84 = - 3
This indicates the sequence is arithmetic with nth term
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 87 and d = - 3 , then
= 87 - 3(n - 1) = 87 - 3n + 3 = 90 - 3n
Answer:
969.70
Step-by-step explanation:
According to the rule of BODMAS multiplication goes first.
B - brackets
O - operations
D - division
M - multiplication
A - addition
S - subtraction
to answer the question, 4 ^ x goes first. which equals 4x. the new equation would be:
4x + 3 = 18
now you have to move +3 over the equals sign which becomes negative (-):
4x = 18 - 3
4x = 15
next you have to get the x to be alone, move the 4 over the equals sign. 15 is now divided by 4:
x = 15/4
x = 3.75