Question

How many different numbers can
be formed using all of the digits 3, 3,
4, 5, 5, 8?
(A) 720
(B) 240
(C) 420
(D) none of these​

Answer 1

Answer:

(D) none of these

hope it helps u

Answer 2

The  numbers can  be formed using all of the digits 3, 3,4,5,5,8 is option(D) none of these

Step-by step explanation:

Given:

         $3,3,4,5,5,6$

To find:

Possible of  numbers ?

Formula:            

$n_n=\frac{n!}{n_1!n_2!n_\infty!}$

where,  

No of arrangements of $n$ distinct object = $n!$

No of arrangements of $n$ objects  where$n_1$  are same ,$n_2$  are same$n_n$  are same.  

Possible of numbers:  

Total no of  digit($n$)=$6$

Repeated digits,  

($n_1$ )=$3-2$ times    

($n_2$)=$5-2$ times  

The values of($n$)  ,($n_1$) and ($n_2$) are substitute to the formula  after we get,    

                                  $n_n=\frac{n!}{n_1!n_2!n_\infty!}$

                                   $n_n=\frac{6!}{2!.2!}$

                                   $n_n=\left(\frac{6\times5\times4\times3\times2\times1)}{\left(2\times1\right)\left(2\times1\right)}\right)$

                                   $n_n=\frac{720}4$

                                   $n_n=180$

Hence ,the possible  of numbers =$180$