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slices.py
```def first_4(x):
return x[:4]

def first_and_last_4(x):
a=x[:4]
b=x[len(x)-4:]
c=a+b
d="".join(map(str,c))
return d
```

Steven Parker - Will do, thanks for the heads up.

Task 2 is: `OK, this second one should be pretty similar to the first. Make a new function named first_and_last_4. It'll accept a single iterable but, this time, it'll return the first four and last four items as a single value.`

So the way we can achieve this is by using a cool feature of slices. So what we can do is ask for x amount of items from the end of the list.

So for example:

```# Let's say we have this list here in the variable x
x = [1, 2, 3, 4, 5, 6, 7, 8]
# We can use a negative number to get the numbers from the end of the list.
x[-2:]
#Would return
[7, 8]
```

So going back to what we want to do for this question is simply get the first 4 numbers and the last four. Then return a single list with those 8 numbers.

```def first_and_last_4(x):
#To get the first four we can replicate the code created in the first task or call that function.
first_four = x[:4]
# Then to get the last four we simply just use a negative four to get them
last_four = [-4:]
# Lastly we return it.
return first_four + last_four
```

An even cleaner solution would be:

```def first_and_last_4(x):
return first_4(x) + x[-4:]
```

This code will pass the challenge:

```def first_4(x):
return x[:4]

def first_and_last_4(x):
return first_4(x) + x[-4:]
```

My issue with this challenge was returning the last 4 in reverse order. I was able to pass doing:

```def first_and_last_4(list):
new_result = list[:4] + list[-4:]
return new_result
```