This problem can be solved using dynamic programming. Here are the steps:

1. Initialize a dp array of size V+1 with all elements as 0. The dp[i] will store the number of ways we can make the sum i using the given coins. dp[0] is 1 because there is one way to make the sum 0 i.e., by selecting no coins.

2. Iterate over each coin. For each coin, iterate from 1 to freq[i] (both inclusive) and for each j, iterate from V to coin[i]*j (both inclusive). For each iteration, add dp[V - coin[i]*j] to dp[V] and take modulo 1000000007 to avoid overflow.

3. The final answer will be dp[V].

Here is the Python code for the above steps:

```python
def countWays(N, coins, freq, V):
mod = 1000000007
dp = [0]*(V+1)
dp[0] = 1
for i in range(N):
for j in range(1, freq[i]+1):
for v in range(V, coins[i]*j-1, -1):
dp[v] = (dp[v] + dp[v - coins[i]*j]) % mod
return dp[V]
```

In the given example, N = 3, coins = [1, 2, 3], freq = [1, 1, 3], V = 6. The output of countWays(N, coins, freq, V) will be 2.

Question

This problem can be solved using dynamic programming. Here are the steps:

1. Initialize a dp array of size V+1 with all elements as 0. The dp[i] will store the number of ways we can make the sum i using the given coins. dp[0] is 1 because there is one way to make the sum 0 i.e., by selecting no coins.

2. Iterate over each coin. For each coin, iterate from 1 to freq[i] (both inclusive) and for each j, iterate from V to coin[i]*j (both inclusive). For each iteration, add dp[V - coin[i]*j] to dp[V] and take modulo 1000000007 to avoid overflow.

3. The final answer will be dp[V].

Here is the Python code for the above steps:

```python
def countWays(N, coins, freq, V):
    mod = 1000000007
    dp = [0]*(V+1)
    dp[0] = 1
    for i in range(N):
        for j in range(1, freq[i]+1):
            for v in range(V, coins[i]*j-1, -1):
                dp[v] = (dp[v] + dp[v - coins[i]*j]) % mod
    return dp[V]
```

In the given example, N = 3, coins = [1, 2, 3], freq = [1, 1, 3], V = 6. The output of countWays(N, coins, freq, V) will be 2.

Knowee AI · Accepted Answer

This problem can be solved using dynamic programming. Here are the steps:

1. Initialize a dp array of size V+1 with all elements as 0. The dp[i] will store the number of ways we can make the sum i using the given coins. dp[0] is 1 because there is one way to make the sum 0 i.e., by selecting no coins.

2. Iterate over each coin. For each coin, iterate from 1 to freq[i] (both inclusive) and for each j, iterate from V to coin[i]*j (both inclusive). For each iteration, add dp[V - coin[i]*j] to dp[V] and take modulo 1000000007 to avoid overflow.

3. The final answer will be dp[V].

Here is the Python code for the above steps:

```python
def countWays(N, coins, freq, V):
    mod = 1000000007
    dp = [0]*(V+1)
    dp[0] = 1
    for i in range(N):
        for j in range(1, freq[i]+1):
            for v in range(V, coins[i]*j-1, -1):
                dp[v] = (dp[v] + dp[v - coins[i]*j]) % mod
    return dp[V]
```

In the given example, N = 3, coins = [1, 2, 3], freq = [1, 1, 3], V = 6. The output of countWays(N, coins, freq, V) will be 2.

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