Maximizing Possibilities: Solving Four-Letter Word Order Question

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SUMMARY

The discussion focuses on calculating the number of four-letter words that can be formed in increasing order from the English alphabet. The initial formula proposed, 1 + (3 x (25-n)), was deemed ineffective. The correct approach involves using nested summations to count the valid combinations, represented as P=\sum_{k=1}^{25}\sum_{l=k}^{25}\sum_{m=l}^{25}\sum_{n=m}^{25}1. This method accurately accounts for the constraints of letter selection based on their order.

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Question: How many four letter words can be made in increasing order. For instance, "aade" is a possibility, but "adae" is not.

Solution: I honestly don't understand how I can solve this, I tried making a formula but it doesn't seem to work. The formula I came up with is: 1 + (3 x (25-n)) where n = 0 when the alphabet is a, n = 1 when the alphabet is b, and so forth. Any help would be greatly appreciated since I have to hand in the solution to this problem by tomorrow morning's class.



Thanks.
 
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Your first letter can be anyone between 1 and 25; call it k.
We have 1<=k<=25
Your second letter can be anyone between k and 25, call it l
We have k<=l<=25
Your third letter can be anyone between l and 25, call it m.
We have l<=m<=25
Your fourth letter can be anyone between m and 25; call it n
We have m<=n<=25.

Thus, counting up all possibilities we've got (call that number P), we may write:
[tex]P=\sum_{k=1}^{25}\sum_{l=k}^{25}\sum_{m=l}^{25}\sum_{n=m}^{25}1[/tex]

What remains is to perform the summation. :smile:
 
Last edited:
Do you need to use the 4 letters a,a,d,e? With the 'a' twice?
In that case, simple counting will do.

Anyhoo, you might as well use numbers instead of letters a,b,c <-> 1,2,3,...
Maybe that's easier to use.
 

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