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Arranging a 6 digit number with 2 identical digits

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Arranging a 6 digit number with 2 identical digits

by gmattesttaker2 » Sat Jan 11, 2014 10:50 am
Hello,

Can you please tell me if my approach is correct:

How many different 6-digit codes can be made by re-arranging all the digits of the number 123,451 if the two 1s cannot be next to each other?

The answer given is 420


I was trying to solve this as follows:

Number of ways in which the two 1's cannot be next to each other = Total arrangements - Number of ways in which the two 1's will be next to each other.

Total arrangements = (6 x 5 x 4 x 3 x 2 x 1)/2! = 6!/2! = 360

To find the number of ways in which the two 1's will be next to each other I group the two 1's as one entity. Hence we now have 5 elements which can be arranged in 5! = 120 ways.

So, Number of ways in which the two 1's cannot be next to each other = 360 - 120 = 240


Thanks for your help,
Sri
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by theCodeToGMAT » Sat Jan 11, 2014 8:00 pm
Sri, I don't see any mistake in your solution. I doubt on the OA

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by gmattesttaker2 » Tue Jan 14, 2014 10:40 pm
theCodeToGMAT wrote:Sri, I don't see any mistake in your solution. I doubt on the OA

What's the source?
Hi Rahul,

I found this question online but the OA given was 420. I guess it was wrong then. Thanks for your help.

Best Regards,
Sri
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by Sankeerthana » Tue Jan 14, 2014 11:12 pm
Hi Sri,

Even I computed 240. OA is wrong.
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by Brent@GMATPrepNow » Wed Jan 15, 2014 7:01 am
gmattesttaker2 wrote:Hello,

Can you please tell me if my approach is correct:

How many different 6-digit codes can be made by re-arranging all the digits of the number 123,451 if the two 1s cannot be next to each other?

The answer given is 420


I was trying to solve this as follows:

Number of ways in which the two 1's cannot be next to each other = Total arrangements - Number of ways in which the two 1's will be next to each other.

Total arrangements = (6 x 5 x 4 x 3 x 2 x 1)/2! = 6!/2! = 360

To find the number of ways in which the two 1's will be next to each other I group the two 1's as one entity. Hence we now have 5 elements which can be arranged in 5! = 120 ways.

So, Number of ways in which the two 1's cannot be next to each other = 360 - 120 = 240


Thanks for your help,
Sri
Great solutions, Sri.

For those students out there who may be wondering how Sri calculated the part in green above, I thought I'd add some rationale.
---------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical O's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------

In the original questions, we want to arrange the digits 1, 1, 2, 3, 4, and 5
There are 6 objects altogether, and we have 2 identical 1's.
So, the total number of arrangements = 6!/2!
= 360

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by gmattesttaker2 » Wed Jan 15, 2014 6:40 pm
Brent@GMATPrepNow wrote:
gmattesttaker2 wrote:Hello,

Can you please tell me if my approach is correct:

How many different 6-digit codes can be made by re-arranging all the digits of the number 123,451 if the two 1s cannot be next to each other?

The answer given is 420


I was trying to solve this as follows:

Number of ways in which the two 1's cannot be next to each other = Total arrangements - Number of ways in which the two 1's will be next to each other.

Total arrangements = (6 x 5 x 4 x 3 x 2 x 1)/2! = 6!/2! = 360

To find the number of ways in which the two 1's will be next to each other I group the two 1's as one entity. Hence we now have 5 elements which can be arranged in 5! = 120 ways.

So, Number of ways in which the two 1's cannot be next to each other = 360 - 120 = 240


Thanks for your help,
Sri
Great solutions, Sri.

For those students out there who may be wondering how Sri calculated the part in green above, I thought I'd add some rationale.
---------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical O's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------------------------

In the original questions, we want to arrange the digits 1, 1, 2, 3, 4, and 5
There are 6 objects altogether, and we have 2 identical 1's.
So, the total number of arrangements = 6!/2!
= 360

Cheers,
Brent
Hello Brent,

Thanks a lot. In-fact I had your MISSISSIPPI example in mind only when I solved this problem. You had given an excellent explanation of this technique in one of your earlier posts as well. Thanks a lot for your help.

Best Regards,
Sri
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