zzz

# Question 2

## a)
```{r}
qs2a = function(n){

  s1 = 0                                        # Outer sum
  for (k in 0:n){                               # Outer summation

    s2 = 0                                      # Inner sum
    for (j in 0:k) {                            # Inner summation
      s2 = s2 + ((-1)^(k  - j))*choose(k,j)*j^n # Formula applied
    }

    s1 = s1 + s2/factorial(k)                   # Multiplying the inner sum by 1/k!
  }
  return(s1)                                    # Returning the total sum
}
```

###### $b) test case: P_5$
```{r}
qs2a(5)                       # Finding P5
```

###### $a) test case: P_15$
```{r}
qs2a(15)                      # Finding P15
```

## b)
```{r}
qs2b = function(n){
  s = 0                                   # Outer sum
  for (k in 0:n){                         # Outer summation

    # Changed the inner sum to a vector replacing j with 0:k for single evaluation of inner summation
    s = s + sum(((-1)^(k - 0:k))*choose(k,0:k)*(0:k)^n)/factorial(k)
  }
  return(s)                               # Returning total sum
}
```

###### $b) test case: P_5$
```{r}
qs2b(5)                                   # Finding P5
```

###### $b) test case: P_15$
```{r}
qs2b(15)                                  # Finding P15
```

## c)
```{r}
qs2c = function(n){
  # Replacing both summations with vectors: k with 0:n and j with
  k = 0:n
  return(sum ( sum(((-1)^(k - 1:k))*choose(k, 1:k )*( 1:k )^n)/factorial(k) ) )
}
```

###### $c) test case: P_5$
```{r}
qs2c(5)                                   # Finding P5
```

###### $c) test case: P_15$
```{r}
qs2c(15)                                  # Finding P15
```