How many different positive integers having six digits are

Question

How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A) 360
B) 720
C) 840
D) 1,080
E) 1,440

A) 360
B) 720
C) 840
D) 1,080
E) 1,440

Bunuel · Accepted Answer

3445 78  --> permutations of these 6 digits where 2 of them are identical (two 4's) is 6!/2! = 360;

3445 77  --> permutations of these 6 digits where 2 of them are identical (two 4's) and 2 others are also identical (two 7's) is 6!/(2!2!) = 180;

3445 88  --> permutations of these 6 digits where 2 of them are identical (two 4's) and 2 others are also identical (two 8's) is 6!/(2!2!) = 180.

Total = 360 + 180 + 180 = 720.

Answer: B.

Hope it's clear.

suresh8055 · Answer

Bunuel can you please explain in deep ? I am not getting this :cry:

Rushil · Answer

Hey Suresh!

As per my understanding, the concept is about simple arrangement.
Since the question states that each of the last digits are a 7 or an 8 and that there are only two places left at the end.
This gets us to 3 cases: Case 1 - a 7 and then an 8 in the end
 Case 2 - a 7 and a 7 in the end
 Case 3 -an 8 and an 8 in the end

Now, for each of these cases a random selection of the numbers have been taken because they have to be arranged anyways.
So for Case 1 - We have 344578. As per the permutations concept, since there are repeating items that have to be permuted we divide the number by the frequency of the repeating item.
So it results as 6!/2! (4 repeats twice)

For Case 2 - We have 344577. As explained for Case 1, we have 6!/(2!*2!). Here both 4 and 7 repeat twice.

Similarly, in Case we have 344588. A permutation results 6!/(2!*2!).

The grand total on all the 3 cases is 6!/2!+ 6!/(2!*2!)+6!/(2!*2!) = 360+180+180 = 720

Regards
Rushil

EMPOWERgmatRichC · Answer

Hi All,

For this question, there are a couple of different ways to "do the math." Here's a way that breaks down the possibilities into small groups:

Based on the prompt, we know we have: 
3, 4, 4, 5, (7 or 8), (7 or 8)

This breaks down into 3 possibilities: 
3, 4, 4, 5, 7, 7 
3, 4, 4, 5, 7, 8 
3, 4, 4, 5, 8, 8

We can now calculate how many 6-digit numbers there are for each possibility.

If we had 6 DIFFERENT numbers (e.g. 1, 2, 3, 4, 5, 6), there would be 6! = 720 different 6-digit numbers

For 3, 4, 4, 5, 7, 7 though, we have some "duplicate numbers" (two 4s and two 7s), which affect the math. Each of those "sets of 2" means that we have to divide by 2!

So, here we'd have 6!/{2!2!] = 720/4 = 180 different 6-digit numbers

For 3, 4, 4, 5, 8, 8 we have the same "duplicate numbers" situation (two 4s and two 8s)….

So, here we'd also have 6!/  = 180 different 6-digit numbers

For 3, 4, 4, 5, 7, 8 we have just one group of duplicates (two 4s), so we divide by just 2!….

Here, we'd have 6!/2! = 720/2 = 360 different 6-digit numbers

In total, we have 180 + 180 + 360 = 720 different 6-digit numbers

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

ScottTargetTestPrep · Answer

We need to determine the number of 6-digit numbers in which one of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8.

Option #1: The number has two 7s

3-4-4-5-7-7

We use the indistinguishable permutations formula to determine the number of ways to arrange 3-4-4-5-7-7:

6!/(2! x 2!) = (6 x 5 x 4 x 3 x 2)/(2 x 2) = 6 x 5 x 3 x 2 = 180 ways

Option #2: The number has two 8s

3-4-4-5-8-8

6!/(2! x 2!) = 180 ways

Option #3: The number has a 7 and an 8

3-4-4-5-7-8

6!/2! = (6 x 5 x 4 x 3 x 2)/2 = 6 x 5 x 4 x 3 = 360 ways

Thus, the number can be created in 180 + 180 + 360 = 720 ways.

Answer: B

cyd37 · Answer

How come only 3 cases in the end: 7,7 8,8 and 7,8. How come 8,7 is not considered?

Bunuel · Answer

6!/2! = 360 gives all permutations of 344578:
344578;
344587;
783445;
...

EMPOWERgmatRichC · Answer

Hi cyd37,

We're ultimately looking for three separate groups of numbers:

1) A group that includes two 7s
2) A group that includes two 8s
3) A group that includes one 7 and one 8

The calculations that you refer to already include a group that has one 8 and one 7; it's the first group (re: 344578), which includes the numbers 344,578 and 344,587 among its 360 options. 
 
GMAT assassins aren't born, they're made,
Rich

damd · Answer

-------

the question statement clearly mentions "each of the other digits is a 7 or an 8", and doesn't anywhere mention that "each of the other digits is a 7 or an 8  or BOTH

TarunKumar1234 · Answer

How many different positive integers having six digits are there, where exactly on of the digits is a 3, exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

When two 7s are used = 6!/2!2! = 720/4 =180
when one 7 and one 8 is used = 6!/2! = 360
when to 8s are used = 6!/2!2! = 720/4 =180

So, Total nos. of ways= 180+360+180 = 720

So, I think B.

GraceSCKao · Answer

I am confused with the last sentence--"each of the other digits is a 7 or an 8," which seems that there are only two possibilities, 77 or 88, to me.

Could anyone clarify why there are four possibilities, 78, 87, 77 and 88?

For the latter version, I expect the statement to be more like "for the remaining two digits, the digit is either 7 or 8."

(As a nonnative English speaker, sometimes I find the most difficult part of a quant question is the language, not the formula.....)

Thanks in advance if anyone could help clarify.

EMPOWERgmatRichC · Answer

Hi GraceSCKao, The prompt uses the specific wording ...."and EACH of the other digits is a 7 or an 8." Notice the singular pronoun and verb in the sentence - meaning that we should be thinking of the two other digits as individuals (and not 'restricted' to being the same digit). By extension, the last two digits could be 77, 78, 87 or 88. IF the prompt had used the phrase "BOTH of the digits ARE the same and either a 7 or an 8", then we would have just two options (77 or 88). GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

Vibhatu · Answer

idinuv  please write the question properly, you have written ”where exactly on of the digits is 3 .. it should be  “one of the digits”

Posted from my mobile device

Bunuel · Answer

Fixed the typo. Thank you!

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How many different positive integers having six digits are

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