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Answer:
It can be found using factorials (!). ! denotes that a number will be multiplied by every number between it and 1 inclusive. so for example:
4!=4∗3∗2∗1
As repeats are not allowed, for every digit you choose, there will be one less option for the next digit. Factorials can provide the number of combinations in this instance, as the value is being multiplied by the number of options.
In your question, only the first 5 digits selected contribute to the number of combinations, after 5 it doesn't matter what order the digits come in. If you were to just use 8! it would still account for these numbers. to prevent this, you can divide 8! by the factorial of obsolete digits (3!).
So the answer to your question is 8!/3! = 6720
The following is to help you with future questions, so do not read further if you aren't interested:
What is mentioned above can be used to derive the equation:
n!/(n-r)! = possible combinations
Where n is the number you have to choose from, and r is the number that must be chosen. Note this only works when no repeats are allowed, i.e the number of options following each digit decreases by one.
Alternatively to all of the above, you can just use the nPr function on a scientific calculator;
n = the number of options you have to choose from
r = the number of options you have to choose
P stands for permutation, it is not a value you can change, it just represents the function.
Step-by-step explanation:
don't know kong tama ba correct me nalang kung mali