Letters in the word HULULULU are rearranged. The probability of all three L being together is

Question

Letters in the word HULULULU are rearranged. The probability of all three L being together is (A) (3/20) (B) (2/5) (C) (3/28) (D) (5/23)

Tardigrade Admin · Accepted Answer

Given words, HULULULU ∴ Total number of sample space =(8 !/4 ! 3 !) and total number of ways all three L being together =(6 !/41) Required probability =((6 !/4!)/(8 !)4 ! 3 !) =(6! × 3!/8 !)=(3/28)

Q. Letters in the word HULULULU are rearranged. The probability of all three L being together is

$\frac{3}{20}$

$\frac{2}{5}$

$\frac{3}{28}$

$\frac{5}{23}$

Given words, HULULULU
$∴$ Total number of sample space $= \frac{8 !}{4 ! 3 !}$
and total number of ways all three $L$ being together $= \frac{6 !}{41}$
Required probability $= \frac{\frac{6 !}{4 !}}{\frac{8 !}{4 ! 3 !}}$
$= \frac{6 ! \times 3 !}{8 !} = \frac{3}{28}$

Q. Letters in the word HULULULU are rearranged. The probability of all three L being together is

203​

52​

283​

235​

Given words, HULULULU ∴ Total number of sample space =4!3!8!​ and total number of ways all three L being together =416!​ Required probability =4!3!8!​4!6!​​ =8!6!×3!​=283​

$\frac{3}{20}$

$\frac{2}{5}$

$\frac{3}{28}$

$\frac{5}{23}$

Given words, HULULULU
$∴$ Total number of sample space $= \frac{8 !}{4 ! 3 !}$
and total number of ways all three $L$ being together $= \frac{6 !}{41}$
Required probability $= \frac{\frac{6 !}{4 !}}{\frac{8 !}{4 ! 3 !}}$
$= \frac{6 ! \times 3 !}{8 !} = \frac{3}{28}$