8 Children are standing in a line outside a ticket counter at zoo. 4 of them have a 1 rupee coin each & the remaining four have a 2 rupee coin each. The entry ticket costs 1 rupee each. If all the arrangements of the 8 children are random, the probability that no child will have to wait for a change, if the cashier at the tickets window has no- change to start with is `K`. Then `15K` is equal to

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have 8 children, 4 with 1 rupee coins and 4 with 2 rupee coins. The ticket costs 1 rupee. We need to find the probability that no child will have to wait for change when they buy their tickets.

### Step 2: Define the Conditions
For a child with a 2 rupee coin to buy a ticket without needing change, there must be at least one child with a 1 rupee coin who has already bought a ticket before them. This means that at any point in the line, the number of children with 1 rupee coins must be greater than or equal to the number of children with 2 rupee coins.

### Step 3: Set Up the Problem
Let’s denote the children with 1 rupee coins as A1, A2, A3, A4 and those with 2 rupee coins as B1, B2, B3, B4. We need to arrange these children in such a way that at any point in the arrangement, the number of A's is always greater than or equal to the number of B's.

### Step 4: Use Combinatorics
The total arrangements of 8 children (4 A's and 4 B's) is given by the formula for permutations of multiset:
\[ \text{Total arrangements} = \frac{8!}{4! \cdot 4!} = 70 \]

### Step 5: Count the Favorable Arrangements
To count the favorable arrangements where no child has to wait for change, we can use the concept of Catalan numbers. The number of valid arrangements of 4 A's and 4 B's where at no point the number of B's exceeds the number of A's is given by the 4th Catalan number:
\[ C_n = \frac{1}{n+1} \binom{2n}{n} \]
For \( n = 4 \):
\[ C_4 = \frac{1}{4+1} \binom{8}{4} = \frac{1}{5} \cdot 70 = 14 \]

### Step 6: Calculate the Probability
The probability \( K \) that no child will have to wait for change is given by the ratio of favorable arrangements to total arrangements:
\[ K = \frac{14}{70} = \frac{1}{5} \]

### Step 7: Find \( 15K \)
Now, we calculate \( 15K \):
\[ 15K = 15 \cdot \frac{1}{5} = 3 \]

### Final Answer
Thus, \( 15K = 3 \).

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