Problem can be reduced to = Total Number of arrangements - No of arrangements when alphabets are togethermariah wrote:How many ways are possible to arrange A, O, L, L, and P with two "L" being separated by at least one letter?
Yes but we also need to consider that the two L's can swap among themselves, so we would need to multiply by a factor of 2.piyush_nitt wrote:Problem can be reduced to = Total Number of arrangements - No of arrangements when alphabets are togethermariah wrote:How many ways are possible to arrange A, O, L, L, and P with two "L" being separated by at least one letter?
Number of arrangements without any condition = 5!/2! = 60
Consider LL as one = 4! = 24
=> 60-24
Ans = 36 ??
first, the bad news (for you; good news for the other posters):aroon7 wrote:first lets fill the Ls
L _ L _ _
to fill the first l = 2 options
for second L = 1
total arrangements = 2* 1 * 3 * 2 *1 = 12
now Ls can be:
L _ L _ _
_ L _ L _
_ _ L _ L
L _ _ L _
_ L _ _ L
L _ _ _ L
6 possiblities
so 12 * 6 = 72
OA pls
nope.deep2002 wrote:Wouldn't the possible combinations with no restrictions be 5!/1! since there are 4 alternatives? (A,O,L,P)
I can see that this is right, and I can see that 3! accounts for the permutations of the letters other than L.sureshbala wrote:Folks, this can be answered directly in the following way.
First we can arrange A, O, and P in 3! ways.
Once we arrange them we have 4 places where we have to place the two L's which can be done in 4C2 ways. (remember since two L's are indistinguishable it is 4C2, if they are distinguishable it will be 4P2)
Hence total number of ways is 3! x 4C2 =36
cjb wrote:i don't see an immediate conceptual interpretation. if there is one, perhaps the previous poster might be so kind as to fill us in.sureshbala wrote:How do I "see" that 4C2 is the number of possibilites for the Ls? I can see that it would be 5C2 if it weren't for the rule against them being adjacent. Why is it 4C2 once that rule is in place?
there's a general rule that Nc2 + N = (N + 1)c2. if you want to conceptualize this rule, you can think of the fact that the "original" Nc2 combinations are still there when you add in the (N + 1)th element, but now there are also N more combinations of two things: the new (N + 1)th element, paired with each of the original N elements.
that's the genesis of the 4c2 here: rearrange the equation to (N + 1)c2 - N = Nc2. in this case, that's 5c2 (total unrestricted choices) minus 4 (number of ways to pick 2 adjoining slots) = 4c2.
you could also just memorize the fact that the number of ways to pick 2 slots from a total of N slots, in such a way that they don't adjoin each other, is (N - 1)c2.
as with most obscure memorized facts, it's unlikely that you'll get the chance to use this one. however, if you do get that chance, it could quite well render a "hard" problem very easy indeed.
Dear cjb,cjb wrote:I can see that this is right, and I can see that 3! accounts for the permutations of the letters other than L.sureshbala wrote:Folks, this can be answered directly in the following way.
First we can arrange A, O, and P in 3! ways.
Once we arrange them we have 4 places where we have to place the two L's which can be done in 4C2 ways. (remember since two L's are indistinguishable it is 4C2, if they are distinguishable it will be 4P2)
Hence total number of ways is 3! x 4C2 =36
How do I "see" that 4C2 is the number of possibilites for the Ls? I can see that it would be 5C2 if it weren't for the rule against them being adjacent. Why is it 4C2 once that rule is in place?
That makes sense, thanks.sureshbala wrote:After you place A O and P there will be two places between A O and P and two places at either ends. So you can place 2 L's at any of these 4 places so that no two L's will be together.
