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?
I think 1800
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!
thank you ^^
@dumbcow: "For the "1? ? 2" grouping there are *5 choose 2* possible groups to go in between" why \( \binom{5}{2} \) and not \( ^5P_2 \)?
I am getting \(40×3!\), consistent to \( 4!× ^5P_2×2! \) for the last case.
LOL its ok foolformath, i have figured out how to do this question but nevertheless your help is greatly appreciated!
Wait, I want to know if I am wrong or not.
its actually 4!× ^5P_2×2! x2!
From where do you get the additional \( 2! \) ?
ok that came out wierd but basically is what you had x 2!
because you can interchange the 1 and 2
but then how you explain the other 2! ?
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!
Exactly, dumbcow's answer needs another \(2! \) to be correct.
GOOD detective skills, i didn't even notice that hehehe!
Lol, thanks :)
Ahh yes i missed that...thanks for correcting me
anytime ^.^
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