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We have 4 times C and 2 times V.
Repetition is allowed.
That sums up the possibilities:
C^4*V^2 with C=21 and V=5
= 4,862,025
A fully random 6-letter-password would be almost 64 times stronger.

A team of six students traveled to a mathematics competition by bicycle. On leaving, they distractedly picked up their six helmets at random (each choice of helmet is equally likely). What is the probability that exactly three of the students picked up the correct helmet?
can you explain why? and show the work??

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14y ago
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Wiki User

15y ago

If the consonants are not allowed to be repeated, that is DAD is invalid, but vowels are repeatable, there are 3,591,000 combinations. If vowels cannot be repeated, but consonants can, there are 3,889,620 combinations. If both can be repeated, there are 4,862,025 combinations. And if neither is repeatable, there are 2,872,800 combinations.

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2008_irfan Khan

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4mo ago

10,000

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Q: A phonetic password generator picks two segments randomly for each six letter password. form is CVC. what is the total password population?
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