My idea was as follows, I'm not exactly great in these kind of questions so it would be nice if someone commented.
We got 2 numbers to place.
1st one can be placed in one of the 6 spots: _ X _ X _ X _ X _ X _ (6 spots = 6 underscores), we can do it in 6 different ways
2nd number can be placed in one of the 7 spots (after having placed the first number we got a 6 digit now: _ X _ X _ X _ X _ X _ X _ : 7 spots = 7 underscores), respectively 7 ways.
As a result we get 6*7 = 42 ways
Since we got duplicates coz of numbers being equal, we'll have to cut it in half, which gives us 42/2 = 21, the answer C.

Should be 21.

there are two possibilities for placing 2 3s .

case 1: two 3s were missed consecutively. i.e. he typed 33 and it came blank on screen.
-5-2-1-1-5- in this arrangement we can fit 33 in 6 ways . (Six dashes, each dash represent one possible place for placing 33)

case 2: two 3s are not together, i.e. they have one or more digits between them .
-5-2-1-1-5- , in this arrangement
if we place first 3 at first dash i.e. 35-2-1-1-5- then the other 3 can fit into 5 places.
if we place first 3 at second dash i.e. -532-1-1-5- then the other 3 can fit into 4 places.
if we place first 3 at third dash i.e. -5-231-1-5- then the other 3 can fit into 3 places.
if we place first 3 at fourth dash i.e. -5-2-131-5- then the other 3 can fit into 2 places.
if we place first 3 at Fifth dash i.e. -5-2-1-135- then the other 3 can fit into 1 place.
so total 15 ways.

case 2 + case 1 = 6+ 15 = 21 ways

Answer C

The other numbers are fixed so we have to choose 2 spaces for the 2 numbers.

7C2=21


Answer : C

I just interpreted the question in a simple fashion : How to place 2 in a 7 digit number and how many ways can we do it?

Simple 7C2 = 21 ways.
Hence C

7 places, 2 choices.

7! / 2!5!

7*6*5*4*3*2*1 / 2*1*5*4*3*2*1 = 7*6 / 2*1 = 21.

We can choose ways to place two 3 three in a 7 digit integer for which 5 digits are fixed is by 7C2 ways.
7C2 = 21
Hence answer is C

two ways of fixing two 3's..
both together- any six places-6 ways..
both separately- any two places out of available six places- 6C2=15..
total 21 ways.. C

7C2 will not stand for all Q of these types..
example if instead of two 3's , say it were 3 and 7.. the ans would be 7P2, as order would matter...
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MAGOOSH OFFICIAL SOLUTION:

First of all, notice that the numerals 5-2-1-1-5 must be in that order --- the order may be broken at any point but one or two 3's, but ignoring the 3's, those five must have that order.

As is often the case, in this counting problem we have a variety of ways to frame the question. I believe the most straightforward is as follows. Consider the seven "blank spaces" into which we will write the seven digits of the original number.

_ _ _ _ _ _ _

Now, select any two of those spaces, and put the two 3's there. The two 3's could be next to each other or apart. Suppose, for example, we put them here:

_ _ 3 _ _3 _

At this point, notice that everything else about the number is determined: we simply will put the digits 5-2-1-1-2 in order in the remaining spaces. Just picking two of the seven spaces for the location of the two 3's is enough to determine the entire original number. Well, there are 7C2 ways of selecting two slots from seven, so 7C2 must be the number of possible original numbers that Sid intended to write.

See this post for various techniques on calculating combinations.

7C2 = 21

So there are 21 possible original numbers that Sid could have intended.

Answer = C
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the trap here is that it is not 6C2, but 7C2 b/c 2 numbers of 3 can stand next to each other.

There are two possible cases to examine here:
case i) the two 3's are adjacent
case ii) the two 3's are NOT adjacent

case i) the two 3's are adjacent
Let's add spaces to where the two 3's can appear: _5_2_1_1_5_
There are 6 possible spaces where the two adjacent 3's can appear
So there are 6 possible ways in which the two adjacent 3's can appear


case ii) the two 3's are NOT adjacent
Let's add spaces to where the two 3's can appear: _5_2_1_1_5_
There are 6 possible spaces where the two non-adjacent 3's can appear
We must choose 2 different spaces
Since the order in which we choose the two spaces does not matter, we can use combinations.
We can choose 2 of the 6 spaces in 6C2 ways
6C2 = (6)(5)/(2)(1) = 15
So there are 15 possible ways in which two non-adjacent 3's can appear

Aside: The video below explains how to quickly calculate combinations (like 6C2) in your head

TOTAL number of possible outcomes = 6 + 15 = 21

Answer: C

Cheers,
Brent


RELATED VIDEO

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Solution:

Let’s place a vertical bar at each end and in between each of the 5 known digits:

|5|2|1|1|5|

If the two 3s must be separated by at least one of the five digits that appear, then any of the vertical bars above can be replaced by a 3. Since there are 6 vertical bars, there are 6C2 = (6 x 5)/2 = 15 ways to pick 2 bars and replace them with a 3. (For example, if each of the first and third bars is replaced by a 3, the seven-digit number is 3523115.)

If the two 3s must be together, then any of the vertical bars above can be replaced by the two 3s. Since there are 6 vertical bars, there are 6 ways to pick 1 bar and replace it with the two 3s. (For example, if the last bar is replaced by two 3’s, the seven-digit number is 5211533.)


Therefore, there are a total of 15 + 6 = 21 different seven-digit numbers Sid could have meant to type.

Answer: C
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