4. Suppose we have the fictional word DALDERFARG. (a) How many ways are there to
ID: 3068566 • Letter: 4
Question
4. Suppose we have the fictional word DALDERFARG. (a) How many ways are there to arrange all of the letters? (b) What is the probability that the 1st letter is the same as the 2nd letter? (c) What is the probability that an arrangement of all of the letters has the 2 Ds next to each other? (d) What is the probability that an arrangement of all of the letters has thc 2 Ds next to each other aud it has the 2 Rs grouped together (not necessarily the Ds and Rs next to cach other)? (e) What is the probability that an arrangement of all the letters has the 2 Ds before the F?Explanation / Answer
4.
(a) Number of ways to arrange all the letters = ( n! / r1! * r2! * .....* rn! )
where, n = Number of letters in the word
r1 = letters of one kind
r2 = letters of another kind and so on till rn
In our letter , n = 10 = Total number of letters
There are 2 A's , 2 D's and 2 R's
So, Number of ways to arrange all the letters = ( 10! / 2! * 2! * 2!) = 453600
(b) Required probability = P(1st letter is same as 2nd letter)
Favourable cases : AA_ _ _ _ _ _ _ _ and RR_ _ _ _ _ _ _ _ and DD_ _ _ _ _ _ _ _
P(1st letter is same as 2nd letter ) = Cases when1st letter is same as 2nd letter / Total cases
Total cases = 453600
Here, first two letters are fixed we just need to arrange the remaining 8 letters
Cases when 1st letter is same as 2nd letter = (8! / ( 2! * 2!) ) * 3 = 30240
Required probability = P(1st letter is same as 2nd letter) = 30240 / 453600 = 1/15 = 0.0666
(c) P(two D's are next to each other)
We can take 2 D's as one unit and remaining 8 letters
In total there are 9 units to be arrange
Favourable cases =( 9! / (2! * 2! )) = 90720
Total cases = 453600
Required probability = P(two D's are next to each other) = 90720 / 453600 = 1/5 = 0.20
(d) Here we take 2 R's as one unit and 2 D's as one unit and remaining 6 units
In total there are 8 units to be arranged
Favourable cases =( 8! / ( 2! )) = 20160
Total cases = 453600
Required probability = 20160 / 453600 = 2/45 = 0.044
(e) Choosing two places= 9C2 and placing 2 D's before F.
In remaining 7 positions we have 7! ways to arrange them.
Favourable cases = 9C2 * 7! = 181440
Total cases = 453600
Required probability = 181440 / 453600 = 2/5 = 0.40
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