The word ALRIGHT contains 7 distinct letters. If the arrangement must begin with "LG", we treat "LG" as fixed at the beginning of the arrangement. This leaves 5 remaining letters: A, R, I, H, T. The number of ways to arrange these 5 letters is given by the factorial of 5: \[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120. \]
Thus, the number of arrangements of the word ALRIGHT that begin with "LG" is \(120\).
We are given the word ALRIGHT, which has 7 distinct letters. We want to find the number of arrangements of all the letters that begin with the letters LG in that order.
Since the first two letters are fixed as L and G, we only need to arrange the remaining 5 letters (A, R, I, H, T) in the remaining 5 positions.
The number of ways to arrange 5 distinct objects in 5 positions is 5! (5 factorial), which is:
5! = 5 × 4 × 3 × 2 × 1 = 120
Therefore, there are 120 arrangements of the letters in ALRIGHT that begin with LG.
Therefore, the answer is 120.
How many possible words can be created from the letters R, A, N, D (with repetition)?