In Morse code, each symbol is either a dot or a dash. How many sequences of 12 s

Question

In Morse code, each symbol is either a dot or a dash. How many sequences of 12 symbols are possible? How many different 6 letter sequences can be made using the first 11 letters of the alphabet? Assume that repetition of letters is allowed. A musician plans to perform 5 selections for a concert. If he can choose from 7 different selections, how many ways can he arrange his program? A signal is made by placing 3 flags, one above the other, on a flag pole. If there are 9 different flags available, how many possible signals can be flown? In a certain lottery, 3 different numbers between 1 and 8 inclusive are drawn. These are the winning numbers. How many different selections are possible? Assume that the order in which the numbers are drawn is unimportant. How many different five -card hands can be dealt from a deck that has no face cards (40 cards altogether)? In a certain lottery, 3 different numbers between 1 and 13 inclusive are drawn at random. These are the winning numbers. If you choose 3 different numbers at random between 1 and 13, what is the probability you will match the winning numbers? Assume that the order of the numbers is unimportant

Explanation / Answer

14) consider 12 symbols as an empty spaces where either dot or dash can come and sit.
   so each space can take 2 values.On total there are 12 spaces so 2*2*... for 12 times so it goes as 2^12.
  

15)Again you can consider empty spaces here but the number of spaces would be six and
   numbers to fill the spaces will be 11.Since the repititon is allowed,each space can take 11 alphabets.
   so it would be 11*11*... 6 times s0 answer is 11^6

16)as there are 7 from which you need to select 5 and order matters so you would choose permutation here. npr =n!/(n-r)!
   7P5 =7!/(75)! = 2520

17)conider 3 empty spaces which has 9 options to fill.Here repition is not allowed so
   first space can take all the possible 9 flags second can take only 8 as one flag is already placed.
   third can take 7 flags as two falgs are placed. so total possible combinations would be 7*8*9=504.

18)no of 3 number sets = 8C3 =56
   no of winning sets of 3 = 3! = 3*2*1 = 6
  
19)combination because the order here is not important and also repition is not allowed
   as there are 40 from which 5 needs to be selected.it would be 40C5=40!/(40-5)! *5!
   =658008

20)no of ways to win is 1 since the order is not important
   no of possible outcomes: 13C3 = 286
   Probability of getting the winning 3 numbers = 1/286

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