The problem can be solved using the concept of combinations in combinatorics. Here's the step-by-step solution:

1. Since each of the five chests must contain at least one ring, let's first place one ring in each chest. This leaves us with 45 rings to distribute.

2. Now, the problem is reduced to distributing these 45 rings into 5 chests, where a chest can contain any number of rings, including zero.

3. This is a classic problem of distributing n identical items into r different groups, which is given by the formula (n+r-1)C(r-1).

4. Substituting n=45 (remaining rings) and r=5 (chests), we get the number of ways as (45+5-1)C(5-1) = 49C4.

So, the king can store the rings among the chests in 49C4 ways, ensuring that every chest contains at least one ring.

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The problem can be solved using the concept of combinations in combinatorics. Here's the step-by-step solution:

1. Since each of the five chests must contain at least one ring, let's first place one ring in each chest. This leaves us with 45 rings to distribute.

2. Now, the problem is reduced to distributing these 45 rings into 5 chests, where a chest can contain any number of rings, including zero.

3. This is a classic problem of distributing n identical items into r different groups, which is given by the formula (n+r-1)C(r-1).

4. Substituting n=45 (remaining rings) and r=5 (chests), we get the number of ways as (45+5-1)C(5-1) = 49C4.

So, the king can store the rings among the chests in 49C4 ways, ensuring that every chest contains at least one ring.

Knowee AI · Accepted Answer

The problem can be solved using the concept of combinations in combinatorics. Here's the step-by-step solution:

1. Since each of the five chests must contain at least one ring, let's first place one ring in each chest. This leaves us with 45 rings to distribute.

2. Now, the problem is reduced to distributing these 45 rings into 5 chests, where a chest can contain any number of rings, including zero.

3. This is a classic problem of distributing n identical items into r different groups, which is given by the formula (n+r-1)C(r-1).

4. Substituting n=45 (remaining rings) and r=5 (chests), we get the number of ways as (45+5-1)C(5-1) = 49C4.

So, the king can store the rings among the chests in 49C4 ways, ensuring that every chest contains at least one ring.

A king has 50 identical diamond rings, which he stores in five different coloured chests. In how many ways can the king store the rings among the chests such that every chest contains at least one ring?49C450C449C550C5

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