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Re: GRE Math Challenge #16 - An Electrical Engineering website [#permalink]
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atleast four means it can be 4,5,6...........,

so
no of ways arranging four letters from 6 = 6c4 *4! =360 (here 4! beacause we can arrange four different letters in 4! ways)

no of ways arranging five letters from 6 = 6c5 *5! =720

no of ways arranging six letters from 6 = 6c4 *6! =720

total ways are 1800
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Re: GRE Math Challenge #16 - An Electrical Engineering website [#permalink]
what if repetition was allowed ? would the formulae be still used ?
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Re: GRE Math Challenge #16 - An Electrical Engineering website [#permalink]
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fabiha22 wrote:
what if repetition was allowed ? would the formulae be still used ?


when repetitions are allowed process will be different

no of 4 letter password = 6 6 6 6=\(6^4\) (each position can be filled with 6 letters)

no of 5 letter password = 6 6 6 6 6=\(6^5\) (each position can be filled with 6 letters)

no of 6 letter password = 6 6 6 6 6 6=\(6^6\) (each position can be filled with 6 letters)

total number of passwords are =\(6^4\)+\(6^5\) +\(6^6\)
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Re: GRE Math Challenge #16 - An Electrical Engineering website [#permalink]
1800
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Re: GRE Math Challenge #16 - An Electrical Engineering website [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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