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btg practice question -- combinatorial

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btg practice question -- combinatorial

by aleph777 » Sat Feb 12, 2011 5:41 am

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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?

10
16
21
24
27

OA: 21


I'm a bit shaky with this type of problem, but I solved it correctly. I just want to see if my logic was correct:

So we know the total number was supposed to be 7 digits. 5 of those digits are already filled. And the last two are meant to be 3s.

Therefore 7!/5!2! (And that's because 7 is the total number, 5 of those digits don't matter because they're already taken, and the 2 threes are identical, so it doesn't matter which comes first.) Is that the correct way to think about it or did I just get lucky?

Thanks!
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by jaxis » Sat Feb 12, 2011 7:26 am
The way you see it is right.

Another approach:

_5_2_1_1_5_
SO this is the situation here.And the relative positions of already present digits doesnt change.

now we can select 2 places for the missing 3's in 6c2 ways. = 15 ways.

AND

No of cases in which both 3's come togather = 6

33 5 2 1 1 5

5 33 2 1 1 5
.
.
.
5 2 1 1 5 33

Total number of ways 15+6 = 21
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by GMATGuruNY » Sat Feb 12, 2011 12:14 pm
aleph777 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?

10
16
21
24
27

OA: 21


I'm a bit shaky with this type of problem, but I solved it correctly. I just want to see if my logic was correct:

So we know the total number was supposed to be 7 digits. 5 of those digits are already filled. And the last two are meant to be 3s.

Therefore 7!/5!2! (And that's because 7 is the total number, 5 of those digits don't matter because they're already taken, and the 2 threes are identical, so it doesn't matter which comes first.) Is that the correct way to think about it or did I just get lucky?

Thanks!
Since the 5 digits given are fixed, we need to count only the number of options for the two missing 3's. The easiest approach would be to say that there are 7 positions in the number, and the two 3's need to occupy a combination of 2 of these 7 positions.

The number of combinations of 2 that can be made from 7 choices = 7C2 = 21.

The correct answer is C.
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Re: btg practice question -- combinatorial

by Brent@GMATPrepNow » Sun Jul 26, 2020 6:54 am
aleph777 wrote: ↑
Sat Feb 12, 2011 5:41 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?

10
16
21
24
27
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

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Brent
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Re: btg practice question -- combinatorial

by Scott@TargetTestPrep » Sun Aug 02, 2020 8:32 am
aleph777 wrote: ↑
Sat Feb 12, 2011 5:41 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?

10
16
21
24
27

OA: 21


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.

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.

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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