Okay, so I need to figure out how many different values can be obtained from the expression 2·3·4·5 + 1 by inserting parentheses. The original expression without any parentheses is just multiplication first, then addition, which gives 2*3*4*5 +1 = 120 +1 = 121. But the problem says that if we insert parentheses, we can get different values. For example, they showed that inserting parentheses can give 144. So the question is asking how many such different values are possible.

First, I need to recall that parentheses change the order in which operations are performed. Since all the operators here are multiplication and addition, the key is that multiplication is left-associative by default, so when you have 2·3·4·5, it's ((2·3)·4)·5. But by inserting parentheses differently, we can group the numbers in different ways, especially by combining some multiplications and additions in different orders.

But since addition is only at the end, the parentheses can affect where the addition happens. Wait, no. The expression is 2·3·4·5 +1. So the addition is only at the end. So actually, the only way parentheses can affect the outcome is by changing the order of multiplications before the addition. Wait, but if we group some multiplications and additions together, but addition is only once at the end. Hmm, maybe not. Wait, let me think.

Wait, the original expression is 2·3·4·5 +1. So all the multiplications are left-associative, so 2*3=6, 6*4=24, 24*5=120, then +1=121. But if we insert parentheses, say, (2·3·4)·(5 +1). Wait, that's the example given, which is (2·3·4)·(5+1) = 24·6=144. So here, the parentheses are grouping 5+1 together, which is 6, and then multiplying that with 2·3·4=24, so 24*6=144. So in this case, the addition is performed after the multiplication, which changes the result.

But is that the only way? Or can we have other groupings? For example, maybe grouping 2·(3·4·5) +1. But in this case, 3·4·5 is 60, so 2*60=120, then +1=121. So that's the same as original.

Alternatively, if we group 2·(3·4)·(5+1), which is the same as the example, 2*12*6=144. So same result.

Alternatively, maybe grouping 2·((3·4)·(5+1)) which is 2*(12*6)=2*72=144. Same result.

Alternatively, ((2·3)·4)·(5+1)=6*4*6=144. Same.

Alternatively, (2·3)·(4·(5+1))=6*24=144. Same.

Alternatively, 2·(3·(4·(5+1)))=2*(3*24)=2*72=144. Same.

So in all cases where the addition is grouped with the 5, we get 144. But if we don't group the addition with the 5, but instead group the 2,3,4,5 in some way, then we get 121. So that's two different values. But maybe there are more.

Wait, but can we get other values? Let's see. Let's think about all possible ways to parenthesize the expression 2·3·4·5 +1. The expression has four multiplication operations and one addition operation. Wait, no. The original expression is 2·3·4·5 +1. So it's a sequence of multiplications followed by an addition. The number of operations is: three multiplications (between 2-3, 3-4, 4-5) and one addition (after 5). So the total number of operations is 4, with three being multiplications and one being addition.

When we insert parentheses, we can change the order in which these operations are performed. However, since addition is only at the end, the parentheses can either group some multiplications together (which will affect the order of multiplications) or group some multiplications with the addition. Wait, but grouping a multiplication with the addition would require that the addition is performed before that multiplication. But in the original expression, the addition is at the end. So if we group a multiplication with the addition, we have to perform that addition first, which would change the order.

Wait, but the original expression is 2·3·4·5 +1. The addition is at the very end. So any parentheses that involve the addition must enclose the 1 and the preceding operations. For example, (2·3·4·5) +1, which is 120 +1=121. Or (2·3·4)·(5 +1), which is 24·6=144. Also, (2·3)·(4·5) +1=6·20 +1=120 +1=121. Wait, that's the same as original. So that's grouping the first two multiplications and the last two multiplications, then adding. So that's 6*20=120, then +1=121.

Alternatively, 2·(3·4·5) +1=2*60 +1=121. So that's same.

Alternatively, if we group 3·(4·5) +1, but that's 3*20 +1=61 +1=62. Wait, but that's not possible because in the original expression, the multiplication is left-associative. Wait, but can we group 3·(4·5) but then where does the 2· come in? Wait, the expression is 2·3·4·5 +1. So if we group 3·4·5 together first, which is 60, then 2·60 +1=121. Alternatively, group 3·(4·5) as part of the 2·3·4·5. Wait, maybe not. Let me think.

Wait, the expression is 2·3·4·5 +1. If we want to group 3·(4·5) first, we have 2·(3·(4·5)) +1=2·60 +1=121. Alternatively, if we group (2·3)·(4·5) +1=6·20 +1=121. So same result.

Alternatively, if we group 2·3·(4·5) +1=6·20 +1=121. So same.

Alternatively, group 2·(3·4)·5 +1=6·5 +1=30 +1=31. Wait, is that possible? Let's check. The original expression is 2·3·4·5 +1. If we group 3·4 first, then 2·12·5 +1=24·5 +1=120 +1=121. Wait, that's same as original. Wait, but if we group (3·4)·5 first, then it's 12·5=60, then 2·60 +1=121. So same.

Alternatively, group 2·(3·4·5) +1=2·60 +1=121. Same.

Alternatively, group (2·3)·4·5 +1=6·4·5 +1=24·5 +1=121. Same.

Alternatively, group 2·3·4·5 +1=120 +1=121. Same.

Alternatively, group 3·(4·5) +1=3·20 +1=60 +1=61. Wait, but how would that work? Because the 2· is still there. Wait, if we group 3·(4·5) +1, but we have to include the 2· somewhere. So maybe 2·(3·(4·5)) +1=2·60 +1=121. So that's same.

Alternatively, group (2·3)·(4·5) +1=6·20 +1=121. Same.

Alternatively, group 2·3·4·5 +1=121. Same.

Wait, so all the groupings where the addition is either at the end or grouped with the last multiplication give 121 or 144. But when I tried grouping 3·4·5 first and then adding 2, it's still 2·60 +1=121. So is 144 the only other value? But the example given is 144, but maybe there's another way to group that gives a different result.

Wait, let me think again. Maybe if we group 2·3·4 first, then (2·3·4)·5 +1=24·5 +1=120 +1=121. Or, group 2·3 first, then 2·3·4·5 +1=6·4·5 +1=24·5 +1=121. Alternatively, group 3·4 first, then 2·3·4·5 +1=2·12·5 +1=24·5 +1=121. Alternatively, group 2·3·4·5 first, then +1=120 +1=121. So all groupings where the addition is either at the end or grouped with the last multiplication give 121. The only other grouping that gives a different result is when we group 5+1 together, which is (2·3·4)·(5 +1)=24·6=144. So that's one different value. But is there another way to group the parentheses to get a different value?

Wait, let's try grouping 2·(3·4·5 +1). But is that allowed? Because the original expression is 2·3·4·5 +1. If we group (3·4·5 +1), that would be changing the order of operations by performing the addition before the last multiplication. But the original expression is 2·3·4·5 +1. So if we group (3·4·5 +1), that would be 3·4·5 +1 first, then multiply by 2. But according to the order of operations, multiplication has higher precedence than addition. However, by inserting parentheses, we can change the order. So 2·(3·4·5 +1) would be 2 multiplied by (3·4·5 +1). Let's compute that: 3·4·5=60, 60 +1=61, then 2·61=122. So that's another value: 122. So that's a third value.

Wait, so that's different from the previous ones. So now I have 121, 122, 144. Maybe there's more.

Wait, let's try another grouping. For example, (2·3)·4·5 +1. But that's same as original. Or 2·(3·(4·5) +1). Let's compute that: 4·5=20, 3·20=60, 60 +1=61, 2·61=122. Same as before. So that's same as the previous one.

Alternatively, group (2·3·4)·(5 +1)=24·6=144. Also, group 2·(3·4·(5 +1))=2·3·4·6= same as 2·3=6, 6·4=24, 24·6=144. Same as before.

Alternatively, group 2·(3·(4 +1))·5. Wait, but 4 +1 is inside the parentheses, which is not allowed because the original expression doesn't have addition except at the end. Wait, no. The original expression is 2·3·4·5 +1. So you can't group 4 +1 unless you add parentheses somewhere else. Wait, but we can only insert parentheses. So if we group 3·(4 +1), but that would require changing the original expression to 2·3·(4 +1)·5 +1, but that's not possible because the original expression doesn't have a +1 in the middle. So we can't group 4 +1 unless we restructure the entire expression.

Wait, no. Let me think. The original expression is 2·3·4·5 +1. So all the multiplications are between the numbers, and the addition is at the end. So we can only insert parentheses around parts of the multiplications or around the addition with some of the multiplications. So for example, grouping 5+1 is allowed, but grouping 4+1 is not, because there's no 4+1 in the original expression. Similarly, grouping 3·4·5 +1 is allowed, but that would require changing the original expression's structure. Wait, no, if we group (3·4·5 +1), then it's 3·4·5 +1 first, then multiply by 2. But that's allowed by inserting parentheses. So 2·(3·4·5 +1)=2·(60 +1)=2·61=122. So that's a different grouping, which gives 122. So that's a new value.

Alternatively, group (2·3)·4·5 +1. That's same as 6·4·5 +1=24·5 +1=121. So that's same as original.

Alternatively, group 2·3·(4·5 +1). Let's compute that: 4·5=20, 20 +1=21, 3·21=63, 2·63=126. So that's another value: 126. So now we have 121, 122, 126, 144. So that's four different values.

Wait, so let's see. Let's list all possible ways to insert parentheses and compute the results.

First, the original expression: 2·3·4·5 +1 = 120 +1=121.

Case 1: Group 5+1 together. So (2·3·4)·(5 +1)=24·6=144.

Case 2: Group 3·4·5 +1 together. Then 2·(3·4·5 +1)=2·61=122.

Case 3: Group 4·5 +1 together. Then 2·3·(4·5 +1)=2·3·21=6·21=126.

Case 4: Group 3·4 +1 together. Then 2·(3·4 +1)·5=2·13·5=26·5=130.

Case 5: Group 2·3 +1 together. Then (2·3 +1)·4·5=7·4·5=28·5=140.

Case 6: Group 2·3·4 +1 together. Then (2·3·4 +1)·5=25·5=125.

Case 7: Group 2·3·4·5 +1 together. That's the original, 121.

Wait, but are these all possible groupings? Let me check.

Alternatively, group 2·(3·4·(5 +1))=2·3·4·6=144. So that's same as case 1.

Alternatively, group (2·3)·(4·(5 +1))=6·24=144. Same as case 1.

Alternatively, group ((2·3)·4)·(5 +1)=24·6=144. Same as case 1.

Alternatively, group (2·(3·4))·(5 +1)=6·6=36. Wait, 3·4=12, 2·12=24, 24·6=144. Same.

Wait, perhaps all groupings that include the addition with the last multiplication result in 144. Wait, but earlier when I grouped 3·4·5 +1, I got 122, and grouping 4·5 +1 gave 126, grouping 3·4 +1 gave 130, grouping 2·3 +1 gave 140, grouping 2·3·4 +1 gave 125. So each time, grouping different parts of the multiplication sequence with the addition gives different results.

So let's list all possible groupings:

1. ((2·3)·4)·5 +1=6·4·5 +1=24·5 +1=121.

2. (2·(3·4))·5 +1=6·5 +1=30 +1=31. Wait, but is this possible? Wait, 3·4=12, 2·12=24, 24·5=120, 120 +1=121. So same as original.

Wait, no, if we group (2·(3·4))·5 +1, then it's 2·12=24, 24·5=120, 120 +1=121. So same as original.

Alternatively, if we group (2·3)·(4·5) +1=6·20 +1=121. Same as original.

Alternatively, group 2·3·4·5 +1=120 +1=121.

Alternatively, group (2·3·4)·(5 +1)=24·6=144.

Alternatively, group (2·3)·4·5 +1=6·4·5 +1=121.

Alternatively, group 2·(3·4·5) +1=2·60 +1=121.

Alternatively, group 2·3·(4·5) +1=6·20 +1=121.

Alternatively, group (2·3·4)·5 +1=24·5 +1=121.

Alternatively, group 2·(3·4·5 +1)=2·61=122.

Alternatively, group (2·3)·4·5 +1=6·4·5 +1=121.

Alternatively, group 2·3·(4·5 +1)=6·21=126.

Alternatively, group 2·(3·4 +1)·5=2·13·5=130.

Alternatively, group (2·3 +1)·4·5=7·4·5=140.

Alternatively, group (2·3·4 +1)·5=25·5=125.

Alternatively, group (2·3·4·5 +1)=121.

Alternatively, group 2·3·4·(5 +1)=2·3·4·6=144.

Alternatively, group (2·3)·(4·(5 +1))=6·24=144.

Alternatively, group ((2·3)·4)·(5 +1)=24·6=144.

Alternatively, group (2·(3·4))·(5 +1)=6·6=36, but 6·6=36, then 36 +1=37. Wait, no, that would be if we group (2·(3·4))·(5 +1)= (6)·6=36, then 36 +1=37. Wait, but that's different. Wait, but in this case, the addition is grouped with the 5, but where does the 1 go? Wait, no. Wait, if we group (2·3·4)·(5 +1), that's 24·6=144. But if we group (2·(3·4))·(5 +1), that's 6·6=36, then 36 +1=37. Wait, but in that case, the addition is after the multiplication of 6·6=36, so 36 +1=37. But is that a valid grouping? Because the original expression is 2·3·4·5 +1. So if we group (2·(3·4))·(5 +1)=6·(5 +1)=6·6=36, then 36 +1=37. Wait, but in the original expression, the addition is at the end. So if we group (5 +1) first, then the rest is multiplied by 6. So 2·3·4 is 24, then 24·(5 +1)=24·6=144. But if we group (2·(3·4))·(5 +1)=6·6=36, then 36 +1=37. Wait, but that's not correct because the original expression is 2·3·4·5 +1. If we group (3·4) first, then 2·(3·4)=6, then (6)·(5 +1)=6·6=36, then 36 +1=37. Wait, but 2·3·4·5 +1 is different from 2·(3·4)·(5 +1) +1. Wait, no. Wait, the original expression is 2·3·4·5 +1. If we group (3·4) first, then 2·(3·4)=6, then 6·5 +1=30 +1=31. Wait, but if we group (2·(3·4))·(5 +1), that would be 6·6=36, but then where does the +1 come in? Because in the original expression, the +1 is after the 5. So if we group (5 +1) first, then we have 2·3·4·6=2·3·24=144. But if we group (2·(3·4))·(5 +1), that's 6·6=36, and then the +1 is outside? Wait, but the original expression is 2·3·4·5 +1. So if we group (5 +1) first, the rest is multiplied by 2·3·4. But if we group (2·(3·4))·(5 +1), that's 6·6=36, but then the expression is 36 +1=37. Wait, but the original expression is 2·3·4·5 +1. So how does grouping (2·(3·4))·(5 +1) make sense? Because the original expression is 2·3·4·5 +1. So if we group (2·(3·4))·(5 +1), that would be 2·3·4·5 +1=121. But if we group (2·(3·4))·(5 +1), that's 2·3·(4·5 +1)=2·3·21=126. Wait, so the order of grouping affects the result.

Wait, maybe I need to clarify the different ways parentheses can be inserted.

Each grouping corresponds to a different binary tree structure for the multiplication and addition operations. Since the addition is only at the end, the different groupings correspond to different ways of associating the multiplications before the addition.

So, to systematically approach this, let's consider all possible ways to insert parentheses around the multiplications, and then compute the result. Since the addition is fixed at the end, we can focus on the different ways to parenthesize the sequence of multiplications before the addition.

The expression without the addition is 2·3·4·5, which can be parenthesized in different ways. Each parenthesization corresponds to a different order of multiplication. The different parenthesizations are determined by the different ways to insert parentheses in the multiplication sequence.

For the multiplication part (2·3·4·5), the different parenthesizations correspond to different binary trees. The number of different parenthesizations for n multiplications is given by the (n-1)th Catalan number. For n=4, the Catalan number is 14. But since we are only considering the different results (due to associativity), some of these might yield the same value.

But since all multiplications are between numbers, the results can vary depending on the grouping. Let's list all possible parenthesizations for 2·3·4·5 and compute their values:

1. ((2·3)·4)·5 = (6·4)·5 = 24·5 = 120
2. (2·(3·4))·5 = (2·12)·5 = 24·5 = 120
3. 2·((3·4)·5) = 2·(12·5) = 2·60 = 120
4. 2·(3·(4·5)) = 2·(3·20) = 2·60 = 120
5. (2·3)·(4·5) = 6·20 = 120
6. (2·(3·4))·5 = same as above, 24·5=120
7. ((2·3)·4)·5 = same as above, 24·5=120

Wait, so all parenthesizations of 2·3·4·5 result in 120. So regardless of how we group the multiplications, the result is 120. Therefore, the only difference comes from grouping the addition with different parts of the multiplications. Therefore, the different values come from grouping the addition with different parts of the multiplication sequence.

Therefore, the different groupings are as follows:

1. Original: 2·3·4·5 +1 = 120 +1=121
2. Group (3·4·5) +1: 2·(3·4·5 +1)=2·(60 +1)=2·61=122
3. Group (4·5) +1: 2·3·(4·5 +1)=2·3·21=6·21=126
4. Group (3·4) +1: 2·(3·4 +1)·5=2·13·5=26·5=130
5. Group (2·3) +1: (2·3 +1)·4·5=7·4·5=28·5=140
6. Group (2·3·4) +1: (2·3·4 +1)·5=25·5=125
7. Group (5 +1) with the rest: (2·3·4·5 +1)=121 (original)

Wait, but grouping (5 +1) is the same as original. So the different groupings that result in different values are:

- Grouping (3·4·5) +1 gives 122
- Grouping (4·5) +1 gives 126
- Grouping (3·4) +1 gives 130
- Grouping (2·3) +1 gives 140
- Grouping (2·3·4) +1 gives 125

Additionally, if we group (2·3·4·5) +1, that's 121.

But wait, there's another way to group: group (2·3·4·5 +1)=121. But if we group (2·3·(4·5 +1))=2·3·21=126, which we already have.

Alternatively, group (2·(3·4·(5 +1)))=2·3·4·6=144. So that's another value: 144.

Similarly, group ((2·3)·4·(5 +1))=6·4·6=144.

Group (2·(3·(4·(5 +1))))=2·3·24=144.

Group (((2·3)·4)·(5 +1))=24·6=144.

So that's another value: 144.

Therefore, in addition to the groupings that give 121, 122, 125, 126, 130, 140, we have 144 as another value.

Is there any other way to group that gives a different value?

Let's check:

Group (2·3·(4 +1))·5. Wait, but 4 +1 is not part of the original expression. The original expression is 2·3·4·5 +1. So you can't group 4 +1 unless you restructure the expression. So this isn't allowed. Similarly, grouping 3 +1 or 2 +1 is not possible because those additions are not present in the original expression.

Therefore, the possible groupings are limited to inserting parentheses around parts of the multiplication sequence and the addition. So the different groupings that result in different values are:

1. Original: 121
2. Group (3·4·5) +1: 122
3. Group (4·5) +1: 126
4. Group (3·4) +1: 130
5. Group (2·3) +1: 140
6. Group (2·3·4) +1: 125
7. Group (2·3·4·5) +1: 121 (original)
8. Group (2·3·4·5 +1): original
9. Group (2·3·(4·5 +1)): 126
10. Group ((2·3)·4·(5 +1)): 144
11. Group (2·(3·4·(5 +1))): 144
12. Group (((2·3)·4)·(5 +1)): 144

Wait, but some of these are duplicates. For example, grouping (2·3·(4·5 +1))=2·3·21=126, which is same as grouping (4·5) +1. Similarly, grouping ((2·3)·4·(5 +1))=6·4·6=144, which is same as grouping (2·3·4)·(5 +1)=24·6=144. So those are same values. So the unique values are:

- 121
- 122
- 125
- 126
- 130
- 140
- 144

So that's seven different values. Wait, but let's check each grouping to make sure we haven't missed any.

1. Original: 121
2. Group (3·4·5) +1: 2·(60 +1)=122
3. Group (4·5) +1: 2·3·(20 +1)=2·3·21=126
4. Group (3·4) +1: 2·(12 +1)·5=2·13·5=130
5. Group (2·3) +1: (6 +1)·4·5=7·4·5=140
6. Group (2·3·4) +1: (24 +1)·5=25·5=125
7. Group (2·3·4·5) +1: 121 (original)
8. Group (2·3·4·5 +1): original
9. Group (2·3·(4·5 +1)): same as 2·3·21=126
10. Group ((2·3)·4·(5 +1)): same as 6·4·6=144
11. Group (2·(3·4·(5 +1))): same as 2·3·4·6=144
12. Group (((2·3)·4)·(5 +1)): same as 24·6=144

So the unique values are 121, 122, 125, 126, 130, 140, 144. That's seven different values.

Wait, but let me verify each one:

1. 121: original
2. 122: 2·(3·4·5 +1)=2·61=122
3. 126: 2·3·(4·5 +1)=2·3·21=126
4. 130: 2·(3·4 +1)·5=2·13·5=130
5. 140: (2·3 +1)·4·5=7·4·5=140
6. 125: (2·3·4 +1)·5=25·5=125
7. 144: (2·3·4)·(5 +1)=24·6=144

Yes, that's seven different values. Let me check if there are any other possible groupings.

For example, grouping 2·(3·(4 +1))·5. But 4 +1 is not part of the original expression. So that's not allowed. Similarly, grouping (2 +1)·3·4·5. But 2 +1 is not part of the original expression. So those are invalid.

Alternatively, grouping 2·(3 +1)·4·5. But again, 3 +1 is not in the original expression. So those are invalid.

Therefore, the only valid groupings are those that group parts of the multiplication sequence with the addition. Therefore, the total number of different values is seven.

Wait, but earlier when I considered grouping (2·3·4)·(5 +1), that's 24·6=144. Similarly, grouping (2·3·(4·5 +1))=2·3·21=126. So that's already included. So seven different values.

Wait, but let me check if there's another way to group. For example, grouping (2·(3·(4 +1)))·5. But again, 4 +1 is not allowed. So no. Similarly, grouping ((2 +1)·3·4·5). But 2 +1 is not allowed. So no.

Therefore, the possible values are 121, 122, 125, 126, 130, 140, 144. That's seven different values.

Wait, but let me check if any of these groupings result in the same value. For example, is there a case where grouping different parts gives the same result? For example, is there a grouping that gives 135? Let's see. Let's check all groupings:

1. 2·3·4·5 +1=121
2. 2·(3·4·5 +1)=2·61=122
3. 2·3·(4·5 +1)=2·3·21=126
4. 2·(3·4 +1)·5=2·13·5=130
5. (2·3 +1)·4·5=7·4·5=140
6. (2·3·4 +1)·5=25·5=125
7. (2·3·4·5 +1)=121
8. (2·3·4·5 +1)=121
9. 2·3·4·(5 +1)=2·3·24=144
10. 2·3·(4·5 +1)=126
11. 2·(3·4·(5 +1))=2·3·4·6=144
12. (2·3·4)·(5 +1)=24·6=144

So the unique values are 121, 122, 125, 126, 130, 140, 144. So that's seven distinct values.

Therefore, the total number of different values is 7.

Wait, but let me check once again to make sure there are no duplicates or missed groupings.

Grouping 1: 121 (original)
Grouping 2: 122
Grouping 3: 126
Grouping 4: 130
Grouping 5: 140
Grouping 6: 125
Grouping 7: 144

Yes, these are all unique. So the answer should be 7 different values.
Therefore, the total number of different values that can be obtained is $\boxed{7}$.
</think>