this would be combination set, since the numbers are the same - i.e. two 3's
C(7,2)=7!/[(7-2)!*2!]=21 additional numbers Sid meant to type with two 3's

artstudent wrote:Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. What appeared instead was the five-digit number 52115. How many different seven-digit numbers could Sid have meant to type?

Why is this not 7!/2!2!2!

Another approach (though I like pemdas' approach more) is to count the spots where they could go.

There are 6 available spots in _5_2_1_1_5_
If the 3s go together, there are 6 places to put them.
If they are separate, there are 6C2 places to put them, = 15.
6 + 15 = 21

Hi Boost,

Can you please explain the below in detail,

If they are separate, there are 6C2 places to put them.
How do we arrive at this?

TIA,
GK

Hi,

When they are separate they can be placed in any of the '_'s in _5_2_1_1_5_ .
There are six such places and we need to select 2. This can be done in 6C2 = 15.

2 cases -- 1) when 3's are together.

this can be done in 6 ways i.e. _5_2_1_1_5_ ( 3352115, 5332115, 5233115, 5213315, 5211335, & 5211533)

2)if the 3's are separated, first can be placed in 6 ways and 2nd can be placed in 5 ways. But, since both the numbers are identical, we have to divide by 2!.

this gives the total as 15.

Hence, the total number of ways = 6+15 = 21