How many ways can you arrange 7 students with specific groupings in a line?

AI Thread Summary
To solve the arrangement of 7 students with specific groupings, treat the two Americans as one unit and the three Indians as another unit, resulting in four units total. The four units can be arranged in 4! (24) ways. Additionally, the two Americans can be arranged among themselves in 2! (2) ways, and the three Indians can be arranged in 3! (6) ways. Therefore, the total number of arrangements is calculated as 4! × 2! × 3! = 288 ways. This method can also be applied if there were only one American and one Indian, simplifying the problem further.
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I have a question here. Please can someone solve this. The question is:
There are 7 students of whom 2 are Americans, 2 are Russians and 3 Indians. hey have to stand in a line so that the two Americans are always together and the three Indians are always together. In how many ways can this be done??
 
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If there were only one American and one Indian, could you solve this problem?
 
In other words, treat the two Americans and the three Indians as being single persons. Once you know how many ways you can line up those "4 people", think about how many new ways you could get by interchanging the Americans only or interchanging the Indians only.
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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