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Sid intended to type a seven-digit number, but the two 3's he meant to

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Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 01 Apr 2015, 04:59
3
7
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

71% (01:11) correct 29% (34:17) wrong based on 160 sessions

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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?

A. 10
B. 16
C. 21
D. 24
E. 27


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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 01 Apr 2015, 05:36
Bunuel 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?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


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
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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 01 Apr 2015, 05:45
Bunuel 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?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


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

7C2=21


Answer : C
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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 01 Apr 2015, 06:00
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Bunuel 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?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


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
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Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 02 Apr 2015, 02:42
1
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.
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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 02 Apr 2015, 02:49
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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.
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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 04 Apr 2015, 20:10
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
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Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 04 Apr 2015, 21:17
1
Bunuel 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?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


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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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 06 Apr 2015, 06:09
1
Bunuel 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?

A. 10
B. 16
C. 21
D. 24
E. 27


Kudos for a correct solution.


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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: Sid intended to type a seven-digit number, but the two 3's he meant to [#permalink]

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New post 27 Nov 2017, 13:59
the trap here is that it is not 6C2, but 7C2 b/c 2 numbers of 3 can stand next to each other.
Re: Sid intended to type a seven-digit number, but the two 3's he meant to   [#permalink] 27 Nov 2017, 13:59
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