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Mathematics 7 Online
OpenStudy (anonymous):

how many numbers of 7 digits can be formed from the digits 1,2,3,4,5,6,7 if there are not more than 2 digits between 1 and 2?

OpenStudy (anonymous):

I think 1800

OpenStudy (dumbcow):

Assuming digits cannot be repeated, look at case where 1 and 2 are next to each other, case where single digit is in between and case where 2 digits are in between. case 1: For the "12" grouping there are 6! ways of arranging them For the "21" grouping there are also 6! ways of arranging them = 2*6! case 2: For the "1?2" grouping there are 5 possible digits to go in between and 5! ways of arranging all the digits...5*5! For the "2?1" grouping there are 5 possible digits to go in between and 5! ways of arranging all the digits...5*5! = 10*5! case 3: For the "1? ? 2" grouping there are 5 choose 2 possible groups to go in between and 4! ways of arranging all the digits...10*4! For the "2? ? 1" grouping there are 5 choose 2 possible groups to go in between and 4! ways of arranging all the digits...10*4! = 20*4! Total numbers is = 2*6! + 10*5! + 20*4!

OpenStudy (anonymous):

thank you ^^

OpenStudy (anonymous):

@dumbcow: "For the "1? ? 2" grouping there are *5 choose 2* possible groups to go in between" why \( \binom{5}{2} \) and not \( ^5P_2 \)?

OpenStudy (anonymous):

I am getting \(40×3!\), consistent to \( 4!× ^5P_2×2! \) for the last case.

OpenStudy (anonymous):

LOL its ok foolformath, i have figured out how to do this question but nevertheless your help is greatly appreciated!

OpenStudy (anonymous):

Wait, I want to know if I am wrong or not.

OpenStudy (anonymous):

its actually 4!× ^5P_2×2! x2!

OpenStudy (anonymous):

From where do you get the additional \( 2! \) ?

OpenStudy (anonymous):

ok that came out wierd but basically is what you had x 2!

OpenStudy (anonymous):

because you can interchange the 1 and 2

OpenStudy (anonymous):

but then how you explain the other 2! ?

OpenStudy (anonymous):

oh because when you have 5c2, these two no.s can swap since there are two no.s. thats why in the case where you only had 1 no. between 1 and 2 you DID NOT have the ADDITIONAL 2!

OpenStudy (anonymous):

Exactly, dumbcow's answer needs another \(2! \) to be correct.

OpenStudy (anonymous):

GOOD detective skills, i didn't even notice that hehehe!

OpenStudy (anonymous):

Lol, thanks :)

OpenStudy (dumbcow):

Ahh yes i missed that...thanks for correcting me

OpenStudy (anonymous):

anytime ^.^

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