If order matters, and there are redundancies, divide by those redundancies. For instance, let’s say I have a 10-letter word, COPYRIGHTS. These letters do not repeat. Therefore, the number of ways I could arrange them is 10!. But with MISSISSIPPI, in which we have 4 ‘S’s, 4 ‘I’s and 2 ‘Ps’, we have to divide as follows: 10!/4!4!2!

Hi Chris,
I arrived at the same answer, but in a different way. Please let me know if my approach is correct.

I feel order does matter here as two 1s placed at different positions lead to a different number. Similarly a 2 placed at different position leads to a different number.

Case 1: Two 1s. Since there are two ones and nineteen 0s to be arranged (order is important), using the MISSISSIPPI rule, we can do it in 21!/(19! * 2!) = 210

Case 2: One 2. Since there is a single 2 and twenty 0s, this can be again achieved by the MISSISSIPPI rule in 21!/(20!) = 21

Thus summing the results of case1 and 2 we get 231.

Hmm…everything seems the same as the explanation except for your assumption that two 1’s placed in different positions lead to different answers. 1001 is the same regardless if I swap the numbers. The reason you got the same answer is you used the combinations formula (which is correct) instead of the permutations formula (which you implied using).

Once I knew to apply the formula, which is the integral part, I saw it as actually 22 integers. Also since between 1 and 21 possible integers (or 0s as shown up there) May not be logically sound, but 22C2 works out without 2nd guessing.

That is a pretty involved problem, and one that is way beyond both the scope of the GMAT and the GRE.

Nonetheless, I am curious – did you have some wonderfully elegant solution? The quickest path I can see the answer requires quite a bit of inelegant grunt work ðŸ™‚

yes! The solution involves combinatorics with repetition and works well (no need for case splitting). I was looking for a canonical formula to generalize it for all sums and integer max min bounds (sum >=9 and lower bound >=1 gets tricky as wrong solutions need to be substracted (combinations as well)

Great! I’m actually going to try to see if I can come up with an elegant approach, when I have more time this week. Fun, challenging twist to the problem! thanks for sharing!

No formula needed… start by building a quick pattern:

10s place = one possibility (2)
100s place = two possibilities (11 and 20)
1,000s place = three possibilities (101, 110 and 200)
10,000s place = four possibilities (1001,1010, 1100 and 2000)
1,000,000,000,000,000,000,000s place = twenty-one possibilities

Now we have a obvious pattern. All we need to do is add up the # of possibilities for each multiple of 10 up to 1,000,000,000,000,000,000,000. To get the answer add 1+2+3+4…+19+20+21 to equal the answer 231

I like this approach as a general way to be successful on the GRE – learn to find patterns. But…a good takeaway from this problem is that the combination formula can be helpful in less obvious contexts, such as the problem above. If you already know the combinations formula, as many do, then this is really helpful.

But you’re right – I see many people thinking – ” oh, I missed this problem because I didn’t know the formula, or didn’t know you could use combinations here. I better remember this formula.” The takeaway is you can’t always know when to apply the correct formula – but you can always experiment.

I appreciate the insight :).

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What is mississipi rule?

If order matters, and there are redundancies, divide by those redundancies. For instance, let’s say I have a 10-letter word, COPYRIGHTS. These letters do not repeat. Therefore, the number of ways I could arrange them is 10!. But with MISSISSIPPI, in which we have 4 ‘S’s, 4 ‘I’s and 2 ‘Ps’, we have to divide as follows: 10!/4!4!2!

Hope that helps!

Hi Chris,

Thanks fo that..it really helped me grasp the concept!

Hi Chris,

I arrived at the same answer, but in a different way. Please let me know if my approach is correct.

I feel order does matter here as two 1s placed at different positions lead to a different number. Similarly a 2 placed at different position leads to a different number.

Case 1: Two 1s. Since there are two ones and nineteen 0s to be arranged (order is important), using the MISSISSIPPI rule, we can do it in 21!/(19! * 2!) = 210

Case 2: One 2. Since there is a single 2 and twenty 0s, this can be again achieved by the MISSISSIPPI rule in 21!/(20!) = 21

Thus summing the results of case1 and 2 we get 231.

Br,Abhijeet

Hmm…everything seems the same as the explanation except for your assumption that two 1’s placed in different positions lead to different answers. 1001 is the same regardless if I swap the numbers. The reason you got the same answer is you used the combinations formula (which is correct) instead of the permutations formula (which you implied using).

Hope that makes sense!

Once I knew to apply the formula, which is the integral part, I saw it as actually 22 integers. Also since between 1 and 21 possible integers (or 0s as shown up there) May not be logically sound, but 22C2 works out without 2nd guessing.

wouldn’t it be just 21C2 = (22*21)/2 = 231

22 digits but last one doesnt add to 2 so we have 21 digits

x1+x2+x3————-+x21=2

21+2-1 C 2 = 22 C 2 = 231.

Hi Rohit789,

I think that is the explanation in the video. Isn’t it?

Hi Chris,

thx! (cases add up as a combinatorial identity wondering if they do for other variations) (i had a typo earlier its 22C2 = 21C2+21C1)

changing the problem sum to 5 instead of 2 everything else same i am getting 53130 as the answer..

Hi Rohit,

I’ll take you word for it :).

That is a pretty involved problem, and one that is way beyond both the scope of the GMAT and the GRE.

Nonetheless, I am curious – did you have some wonderfully elegant solution? The quickest path I can see the answer requires quite a bit of inelegant grunt work ðŸ™‚

yes! The solution involves combinatorics with repetition and works well (no need for case splitting). I was looking for a canonical formula to generalize it for all sums and integer max min bounds (sum >=9 and lower bound >=1 gets tricky as wrong solutions need to be substracted (combinations as well)

Great! I’m actually going to try to see if I can come up with an elegant approach, when I have more time this week. Fun, challenging twist to the problem! thanks for sharing!

No formula needed… start by building a quick pattern:

10s place = one possibility (2)

100s place = two possibilities (11 and 20)

1,000s place = three possibilities (101, 110 and 200)

10,000s place = four possibilities (1001,1010, 1100 and 2000)

1,000,000,000,000,000,000,000s place = twenty-one possibilities

Now we have a obvious pattern. All we need to do is add up the # of possibilities for each multiple of 10 up to 1,000,000,000,000,000,000,000. To get the answer add 1+2+3+4…+19+20+21 to equal the answer 231

Sam,

I like this approach as a general way to be successful on the GRE – learn to find patterns. But…a good takeaway from this problem is that the combination formula can be helpful in less obvious contexts, such as the problem above. If you already know the combinations formula, as many do, then this is really helpful.

But you’re right – I see many people thinking – ” oh, I missed this problem because I didn’t know the formula, or didn’t know you could use combinations here. I better remember this formula.” The takeaway is you can’t always know when to apply the correct formula – but you can always experiment.

I appreciate the insight :).