# 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

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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.

Last updated date: 19th Sep 2023

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