
Letters in the word HULULULU are rearranged. The probability of all three L being together is
A . $\dfrac{3}{{20}}$
B . $\dfrac{2}{5}$
C . $\dfrac{3}{{28}}$
D . $\dfrac{5}{{23}}$
Answer
621.3k+ views
Hint: In order to solve this problem assume the 3 L as one letter then use the permutations and apply the formula of probability that number of favourable outcomes upon total number of outcomes and proceed it further. Doing this you will reach the right answer.
Complete step-by-step answer:
There are 1-H, 4-U and 3-L and hence total 8 alphabets.
Hence the total number of arrangements is 8!.
But there are 4 U and 3L so we will divide 8! with 3! and 4!.
So, the actual number of all arrangements =$\dfrac{{8!}}{{3!4!}}$
Taking all 3-L together, we should treat 3-L as one entity and hence total 6 distinct terms to be arranged. But the arrangement of 3-L =1 as all L are the same.
Now we have only 6 letters in which 4 letters are U.
So, number of arrangements with all 3-L together =$\dfrac{{6!}}{{4!}}$
Hence, the required probability = $\dfrac{{{\text{Number}}\,{\text{of}}\,{\text{favourable}}\,{\text{out}}\,{\text{comes}}}}{{{\text{total number of outcomes}}}}$
=$\dfrac{{\dfrac{{6!}}{{4!}}}}{{\dfrac{{8!}}{{3!4!}}}} = \dfrac{{6!}}{{4!}}{\text{x}}\dfrac{{4!3!}}{{8!}} = \dfrac{{6!{\text{x3!}}}}{{8!}} = \dfrac{{6!{\text{x3!}}}}{{6!{\text{x7x8}}}} = \dfrac{{3{\text{x2}}}}{{56}} = \dfrac{3}{{28}}$
Hence, the correct option is C.
Note: Whenever you face such type of problems you have to assume all the letters as one letter then solve using permutations and obtain the number of arrangements. Here we have asked to find the probability therefore we have obtained the total number of arrangements and the number of arrangements of the asked condition then we have applied the formula of probability. Using this will take you to the right answer.
Complete step-by-step answer:
There are 1-H, 4-U and 3-L and hence total 8 alphabets.
Hence the total number of arrangements is 8!.
But there are 4 U and 3L so we will divide 8! with 3! and 4!.
So, the actual number of all arrangements =$\dfrac{{8!}}{{3!4!}}$
Taking all 3-L together, we should treat 3-L as one entity and hence total 6 distinct terms to be arranged. But the arrangement of 3-L =1 as all L are the same.
Now we have only 6 letters in which 4 letters are U.
So, number of arrangements with all 3-L together =$\dfrac{{6!}}{{4!}}$
Hence, the required probability = $\dfrac{{{\text{Number}}\,{\text{of}}\,{\text{favourable}}\,{\text{out}}\,{\text{comes}}}}{{{\text{total number of outcomes}}}}$
=$\dfrac{{\dfrac{{6!}}{{4!}}}}{{\dfrac{{8!}}{{3!4!}}}} = \dfrac{{6!}}{{4!}}{\text{x}}\dfrac{{4!3!}}{{8!}} = \dfrac{{6!{\text{x3!}}}}{{8!}} = \dfrac{{6!{\text{x3!}}}}{{6!{\text{x7x8}}}} = \dfrac{{3{\text{x2}}}}{{56}} = \dfrac{3}{{28}}$
Hence, the correct option is C.
Note: Whenever you face such type of problems you have to assume all the letters as one letter then solve using permutations and obtain the number of arrangements. Here we have asked to find the probability therefore we have obtained the total number of arrangements and the number of arrangements of the asked condition then we have applied the formula of probability. Using this will take you to the right answer.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

A moving boat is observed from the top of a 150 m high class 10 maths CBSE

