Question

Discrete Math

a.) How many bit strings are there of length five or less, not counting the empty string?

b.) How many different three-letter initials with none of the letters repeated can people have?

c.) How many different three-letter initials are there that begin with the letter B?

d.) How many 5-element DNA sequences end with A?

e.) How many bit strings of length nine both begin and end with 1?

Answer 1

a) number of bit strings of length five or less is 2^5 + 2^4 + 2^3+2^2+2^1 =32+16+8+4+2

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b)first letter can be chosen from 26 lettters

second letters can be chosen from 25 letters

third letter can be chosen from 24 letters

so total = 26*25*24=15,600

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C) first letter is fixed that is B=1

second and third can be choosen from 26 letters

so total =1*26*26= 676

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d) Given five element DNA sequences.

The possible dna bases are A,C,G and T

the fourst four bases can be picked from the 4 letters (A,C,T,G)

and the last letter should be only A

so total = 4*4*4*4*1 = 64

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E) first bt should always be 1 and last bit should always be 1.

From second to eight bit we can choose either 0 or 1

so tota =1*2*2*2*2*2*2*2*1 =2^7 = 128