Solution: 2022 Winter Final - 7

Author: Michiel Smid

Question

A string that is obtained by rearranging the letters of the word \[ \mathrm{ARABICA} \] is stupendous if it contains the substring $\mathrm{CAB}$. For example $\mathrm{ARA\underline{CAB}I}$ and $\mathrm{I\underline{CAB}AAR}$ are both stupendous, but $\mathrm{ARABICA}$ is not. How many stupendous strings are there?

(a)

$70$

(b)

$30$

(c)

$60$

(d)

$40$

(e)

$50$

Solution

Let’s treat CAB as a single letter

That means we have

1 CAB
2 A's
1 R
1 I

Since we have 5 entities in total, it means we have 5 positions

We choose 1 position out of the 5 positions to place CAB: $ \binom{5}{1} = 5 $

We choose 2 positions out of the remaining 4 positions to place A’s: $ \binom{4}{2} = 6 $

We choose 1 position out of the remaining 2 positions to place R: $ \binom{2}{1} = 2 $

We choose 1 position out of the remaining 1 position to place I: $ \binom{1}{1} = 1 $

Thus, the final answer is $ 5 \cdot 6 \cdot 2 \cdot 1 = 60 $