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by artstudent » Sat Aug 13, 2011 12:42 am
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!
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by pemdas » Sat Aug 13, 2011 1:43 am
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!
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by gmatboost » Mon Aug 15, 2011 9:14 am
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
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by GmatKiss » Mon Aug 15, 2011 10:06 pm
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
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by Frankenstein » Mon Aug 15, 2011 10:10 pm
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.
Cheers!

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by saketk » Tue Aug 16, 2011 9:10 am
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
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