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Nov 2020 p51 q7
2741
Four letters are selected from the 10 letters of the word SHOPKEEPER.
Find the number of different selections if the four letters include exactly one P.
Solution
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.
