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Rakesh1987
Bunuel
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s?

A. 55
B. 60
C. 65
D. 70
E. 75
The sequence is like this: 01............10 (19 terms in total)
As two 0s cant be together but two 1s can be lets concentrate on 1s and let use 0 as separator. Our task is to find the number of sequences in which 1s and 11s are separated by 0s.

Minimum number of digit 1 in any sequence is when we use only single 1s and no 11s= (19-1)/2=9. Here there are 9 single 1s.
Maximum number of digit 1 in any sequence is using all pairs of 11s= (19-1)/2=12

As we move from 9 to 12 number of 1, we move from 0 pair to 6 pairs- two in each move why??
Because when we use 10 number of 1- if one pair is used then number of 0s to be used is 9+1=10 and number of 1 is also 10, totals to 20. However, we need to make sequence of 19 1s & 0s. Hence, this is not possible and we will have to move in 2- 11s.

Total number of 1---- Pair of 1,11---- Calculation
9----9,0----1
10----6,2----8!/6!2!=28
11----3,4----7!/3!4!=35
12----0,6----1
Total number of sequences is 1+28+35+1=65. C.

Please hit +1 Kudos if you liked the explanation. :cool:


How did you calculate maximum no. of digit 1 in the sequence, can you please help me with it?
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Rakesh1987
Bunuel
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s?

A. 55
B. 60
C. 65
D. 70
E. 75
The sequence is like this: 01............10 (19 terms in total)
As two 0s cant be together but two 1s can be lets concentrate on 1s and let use 0 as separator. Our task is to find the number of sequences in which 1s and 11s are separated by 0s.

Minimum number of digit 1 in any sequence is when we use only single 1s and no 11s= (19-1)/2=9. Here there are 9 single 1s.
Maximum number of digit 1 in any sequence is using all pairs of 11s= (19-1)/2=12

As we move from 9 to 12 number of 1, we move from 0 pair to 6 pairs- two in each move why??
Because when we use 10 number of 1- if one pair is used then number of 0s to be used is 9+1=10 and number of 1 is also 10, totals to 20. However, we need to make sequence of 19 1s & 0s. Hence, this is not possible and we will have to move in 2- 11s.

Total number of 1---- Pair of 1,11---- Calculation
9----9,0----1
10----6,2----8!/6!2!=28
11----3,4----7!/3!4!=35
12----0,6----1
Total number of sequences is 1+28+35+1=65. C.

Please hit +1 Kudos if you liked the explanation. :cool:

Rakesh1987 Please explain calculation in red color.
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Rakesh1987
Bunuel
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s?

A. 55
B. 60
C. 65
D. 70
E. 75
The sequence is like this: 01............10 (19 terms in total)
As two 0s cant be together but two 1s can be lets concentrate on 1s and let use 0 as separator. Our task is to find the number of sequences in which 1s and 11s are separated by 0s.

Minimum number of digit 1 in any sequence is when we use only single 1s and no 11s= (19-1)/2=9. Here there are 9 single 1s.
Maximum number of digit 1 in any sequence is using all pairs of 11s= (19-1)/2=12

As we move from 9 to 12 number of 1, we move from 0 pair to 6 pairs- two in each move why??
Because when we use 10 number of 1- if one pair is used then number of 0s to be used is 9+1=10 and number of 1 is also 10, totals to 20. However, we need to make sequence of 19 1s & 0s. Hence, this is not possible and we will have to move in 2- 11s.

Total number of 1---- Pair of 1,11---- Calculation
9----9,0----1
10----6,2----8!/6!2!=28
11----3,4----7!/3!4!=35
12----0,6----1
Total number of sequences is 1+28+35+1=65. C.

Please hit +1 Kudos if you liked the explanation. :cool:


How did you calculate maximum no. of digit 1 in the sequence, can you please help me with it?

Just clarified in the original answer itself. Please let me know if it is still not clear; i will elaborate it further.
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Kinshook
Please explain calculation in red color.

The below is the sequence of 19 1s and 0s beginning with 0 and ending with 0, not containing 00 or 111 and is the maximum possible.
0110110110110110110

To make things simple I just deducted 1 from 19 and multiplied it by 2/3. One might ask why?? because the above sequence is exactly this.

I hope it is clear now.
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Bunuel
How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s?

A. 55
B. 60
C. 65
D. 70
E. 75

Asked: How many sequences of 0s and 1s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0s, and contain no three consecutive 1s?

There are strings of 01(x) and 011(y) making first 18 terms, and last term is 0

2x + 3y = 18

If y=0; x =9 => Number of sequences = 1
If y=2; x = 6 => Number of sequences = 8!/6!2! = 28
If y=4; x=3 => Number of sequences = 7!/4!3! = 35
If y=6; x=0 => Number of sequences = 1

Total number of sequences = 1+28+35+1 = 65

IMO C
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