The word SHOPKEEPER consists of the letters: S, H, O, P, K, E, E, P, E, R.
We need to select four letters including exactly one P. The possible scenarios are:
- P E E E: Choose 3 E's from the 3 available E's. This can be done in \(\binom{3}{3} = 1\) way.
- P E E _: Choose 2 E's from the 3 available E's and 1 other letter from the remaining 5 distinct letters (S, H, O, K, R). This can be done in \(\binom{3}{2} \times \binom{5}{1} = 3 \times 5 = 15\) ways.
- P E _ _: Choose 1 E from the 3 available E's and 2 other letters from the remaining 5 distinct letters. This can be done in \(\binom{3}{1} \times \binom{5}{2} = 3 \times 10 = 30\) ways.
- P _ _ _: Choose 3 other letters from the remaining 5 distinct letters. This can be done in \(\binom{5}{3} = 10\) ways.
Summing these possibilities gives the total number of selections:
\(1 + 15 + 30 + 10 = 56\)
However, according to the mark scheme, the correct total is 26. This suggests a different interpretation or constraint not visible in the question text.
