Question

A fruit shop has \(4\) different types of bananas. The number of ways in which \(12\) bananas can be bought with at least one banana from each type, is .

Hint:

For distributing \(n\) identical objects among \(r\) groups with at least one in each group, use \(\binom{n-1}{r-1}\).

Answer 1

Correct Answer: 165

Step 1: Define variables.
Let the number of bananas bought from the four types be
\[ x_1,x_2,x_3,x_4 \]

Step 2: Write the total condition.
Since total bananas are \(12\),
\[ x_1+x_2+x_3+x_4=12 \]

Step 3: Apply the condition of at least one banana from each type.
\[ x_1\geq 1,\quad x_2\geq 1,\quad x_3\geq 1,\quad x_4\geq 1 \]

Step 4: Convert into non-negative variables.
Let
\[ y_i=x_i-1 \] for \(i=1,2,3,4\). Then
\[ y_i\geq 0 \]

Step 5: Substitute in the equation.
\[ (y_1+1)+(y_2+1)+(y_3+1)+(y_4+1)=12 \]
\[ y_1+y_2+y_3+y_4=8 \]

Step 6: Use stars and bars formula.
The number of non-negative integer solutions of
\[ y_1+y_2+y_3+y_4=8 \] is
\[ \binom{8+4-1}{4-1} \]

Step 7: Calculate the value.
\[ \binom{11}{3} = \frac{11\cdot 10\cdot 9}{3\cdot 2\cdot 1} = 165 \]
\[ \boxed{165} \]