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Solving variations with and without repetition

Solving variations with and without repetition


Hello, I have a bit of querying in the probability section. According to the lecture, solving variations with repetition = n power of p, and without repetition = n!/(n-p)! 
Solving two different questions with this approach.
Q1> How many 3 digit number can be made using 5,6,7 only?
A: If repetition allowed : 3^3 = 27, if repetition not allowed: 3!/1! = 6    (Agrees with lecture)
Q2> How many 3 digit number can be made using 0,5,6,7 only?
A: If repetition allowed: 4^3 = 64   (but the actual answer is 48, number starting with zero is not accepted)
If repetition not allowed: 4!/1!=24 (actual answer is 18,since number starting with 0 is not accepted)
Is there any approach with which we can identify and solve such problem using the formula mentioned in the lecture?

1 Answer

365 Team

Hey Sanjesh,
Thanks for reaching out!
In cases like these you have variations where the different positions have different sample spaces. In those cases, you have to split the question in a way, where you can utilize the formulas. 
In this specific case, we can take the first position as a separate event and then use the formulas on the other remaining positions.
If we take the question you suggested, then we have 3 options for the first position and then we use 4^2 for the remaining 2. This equals 3*4^2 = 3 * 16 = 48. If repetition isn’t allowed, we still have 3 options for the first position and then 3!/1! (since we’ve already used 1 of the 4 numbers). Hence, we get 3 * (3!/1!) = 3 * 6 = 18. 
365 Vik

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