I am a chilli

Probability Level 3

You have $\color{#D61F06}{\textbf{20 distinct red chilies}}$ and $\color{#20A900}{\textbf{10 distinct Green chilies}}$ .

The number of $\color{#D61F06}{\text{special red chili powders}}$ you can form is $\color{#D61F06}{R}$ .

The number of $\color{#20A900}{\text{special green chili powders}}$ you can form is $\color{#20A900}{G}$ .

Then the ratio $\dfrac{\color{#D61F06}{R}}{\color{#20A900}{G}}$ is equal to $\dfrac{a}{b}$ where $a$ and $b$ are coprime, positive integers, find $a+b$ .

Details and assumptions -

For making a special chili powder, you can at once use exactly $2$ chilies, and of same color.

$\color{#D61F06}{\text{Red chili}}+ \color{#20A900}{\text{Green chili}} \neq$ any of the asked powders.

$\bullet$ This problem is a part of the set Vegetable combinatorics

The answer is 47.

2 solutions

Tasmeem Reza
Jul 27, 2014

$Number\: of\: \color{#D61F06}{special\: red\: chilli\: powder:\: R} = 20\:choose\:2=\binom{20}{2}$

$Number\: of\: \color{#20A900}{special\: green\: chilli\: powder:\: G} = 10\:choose\:2=\binom{10}{2}$

$Thus\: \frac{\color{#D61F06}{R}}{\color{#20A900}{G}}=\frac{\binom{20}{2}}{\binom{10}{2}}=\frac{38}{9}$

$Thus\: the\: answer\: is\: 38+9=\boxed{47}\: [as\: gcd(38,9)=1]$

$\textbf{Perfectly Colored}$

Aditya Raut - 6 years, 9 months ago

Except the font!

Kartik Sharma - 6 years, 5 months ago

Cody Johnson
Jul 24, 2014

Generating functions.

Where's the equality case??? :D

Dinesh Chavan - 6 years, 10 months ago

I am a chilli

The answer is 47.

2 solutions

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