Placing letters was never a big deal
\[\begin{array} {} \text{Row 1:} & \Huge \Box \Box \Box \Box \\ \text{Row 2:} & \Huge \Box \Box \\ \text{Row 3:} & \Huge \Box \Box \end{array} \]
Let be the number of different ways the letters of the word "JUMPED" can be placed in the boxes shown above so that no row is empty.
Find the value of .
The answer is 72.
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1 solution
The total number of ways to arrange the letters of 'J U M P E D' without any restriction is given by
8 P 6 = 2 ! 8 ! = 2 0 1 6 0
Row 1 cannot be empty ,as row 2 and 3 combined only have 4 spaces between them for the 6 letters. Thus either row 2 or row 3 can be empty ,but this means the other 2 rows will be completely filled.
Thus, we have
2 × 6 ! cases where at least one row is empty.
The number of cases where no row is empty is given by,
K = 2 0 1 6 0 − 2 ∗ 7 2 0 = 1 8 7 2 0
⇒ 2 6 0 K = 2 6 0 1 8 7 2 0 = 7 2 .