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Re: How many eight letter words exist that are composed of Xs an [#permalink]
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hussi9 wrote:
How many eight letter words exist that are composed of Xs and Ys, and which contain neither three consecutive Xs nor three consecutive Ys?

(1) 74 (2) 66 (3) 76 (4) 68 (5) none of these


Whats the source? i see a CAT in this:P
its D .68

I see this more of a pattern question + probability. Using pattern is very useful here as we have a constraint .

let f(n) = no of words with Xs and Ys
f(1) = 2 (XY, YX)
f(2)= 4 (XY XX, YX, YY)

Now the problem is F(3) .. we cant have 3 consecutive X or Y
so we can write F(3) = F(2)+(1).. this will give us a pattern
F(n) = f(n-1)+f(n-2)
f(3) = 2+4 = 6
f(4) = 6+4 = 10
f(5) = 10+6 = 16
f(6) = 16+10 = 26
f(7) = 26+ 16 = 42
f(8) - 42 + 26 = 68
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Re: How many eight letter words exist that are composed of Xs an [#permalink]
sudhir18n wrote:
hussi9 wrote:
How many eight letter words exist that are composed of Xs and Ys, and which contain neither three consecutive Xs nor three consecutive Ys?

(1) 74 (2) 66 (3) 76 (4) 68 (5) none of these


Whats the source? i see a CAT in this:P
its D .68

I see this more of a pattern question + probability. Using pattern is very useful here as we have a constraint .

let f(n) = no of words with Xs and Ys
f(1) = 2 (XY, YX)
f(2)= 4 (XY XX, YX, YY)

Now the problem is F(3) .. we cant have 3 consecutive X or Y
so we can write F(3) = F(2)+(1).. this will give us a pattern
F(n) = f(n-1)+f(n-2)
f(3) = 2+4 = 6
f(4) = 6+4 = 10
f(5) = 10+6 = 16
f(6) = 16+10 = 26
f(7) = 26+ 16 = 42
f(8) - 42 + 26 = 68


Bunuel, could you shed some light over here?

Cheers!
J
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Re: How many eight letter words exist that are composed of Xs an [#permalink]
1
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Expert Reply
jlgdr wrote:
sudhir18n wrote:
hussi9 wrote:
How many eight letter words exist that are composed of Xs and Ys, and which contain neither three consecutive Xs nor three consecutive Ys?

(1) 74 (2) 66 (3) 76 (4) 68 (5) none of these


Whats the source? i see a CAT in this:P
its D .68

I see this more of a pattern question + probability. Using pattern is very useful here as we have a constraint .

let f(n) = no of words with Xs and Ys
f(1) = 2 (XY, YX)
f(2)= 4 (XY XX, YX, YY)

Now the problem is F(3) .. we cant have 3 consecutive X or Y
so we can write F(3) = F(2)+(1).. this will give us a pattern
F(n) = f(n-1)+f(n-2)
f(3) = 2+4 = 6
f(4) = 6+4 = 10
f(5) = 10+6 = 16
f(6) = 16+10 = 26
f(7) = 26+ 16 = 42
f(8) - 42 + 26 = 68


Bunuel, could you shed some light over here?

Cheers!
J


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