interview question
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I would tell the interviewer that it's not a well-formed question. You can draw an infinite number of curves through a series of points, and if you're not looking for the simplest, most obvious one, then any of those infinite possibilities are valid answers.
yes. Maybe you could find pattern inside the series.
That's great. Thanks.every integer is a sum of at most 3 signed squares...the basis set for representing positive integers with positive squares is, so 49 is never used.
How about 52?
every integer is a sum of at most 3 signed squares...the basis set for representing positive integers with positive squares is, so 49 is never used.
That's pretty obscure. I found another obscure pattern you can use: It's the square of prime powers (prime powers being prime numbers raised to whole numbers). This would make x = 7^2 = 49, and y=8^2 = 64.
I stand by my first response that this question is not well-defined and has an infinite number of answers.
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So you're saying you want to draw 6 cards from a deck randomly and get at least 1 of each suit?
(52/52) * (39/51) * (26/50) * (13/49) * (48/48) * (47/47) = 2197/20825 = 0.1055.
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The fact that you can draw an infinite number of curves does not mean there are an infinite number of useful patterns.
yes. your understanding is right.
your method is to draw "one color card 1, another different color card 2, another different color card 3,another different color card 4, another card 5, another card 6"?
Is is possible " one color card 1, another different color card 2, another different color card 3, another card 4, another card 5,another different color card 6"?
This is the solution for getting 4 different suits out of 4 cards. You have to account for getting them in different orders
would you like to explain the reason?
Yes, hence my original post about eliminating the most obvious pattern, which is clearly squares of natural numbers. Once you've eliminated that, there's no clear "useful" pattern.
Sure. You start with getting 6 out of the 52 available cards. Then you have to eliminate the combinations that only appear in 3 suits, but in doing so, you've double eliminated combinations from 2 suits, so you have to add those back. In doing that, you've re-added combinations from 1 suits so you subtract that again.
This was a quick guess so I'm not 100% sure it's right.
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Ugh, you're right, I didn't account for order.
Choose a suit, take 2 cards, choose another suit, take 2 cards, the other two suits have 1 card.
Choose a suit, take 3 cards, the other three suits have 1 card.
((4C1)*(13C2)*(3C1)*(13C2)*(2C1)*(13C1)*(1C1)*(13C1) + (4C1)*(13C3)*(3C1)*(13C1)*(2C1)*(13C1)*(1C1)*(13C1)) / (52C6)
This is my next best guess. I haven't done these problems in a few years.
Choose a suit, take 2 cards, choose another suit, take 2 cards, the other two suits have 1 card.
Choose a suit, take 3 cards, the other three suits have 1 card.
((4C1)*(13C2)*(3C1)*(13C2)*(2C1)*(13C1)*(1C1)*(13C1) + (4C1)*(13C3)*(3C1)*(13C1)*(2C1)*(13C1)*(1C1)*(13C1)) / (52C6)
This is my next best guess. I haven't done these problems in a few years.
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I actually thought darth's answer was very clever.
I'm not saying it's not. My argument is this: are you 100% certain that darth's answer is the one they intended? If not (and I can't see how you can be certain given its obscurity), then the question isn't well-defined.
