How many words can be formed using director such that vowels always stay together?
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This can be solved using permutation.
The three vowels(I, E , O) can be arranged among themselves in 3! (3×2×1)=6 ways.
Next, we can consider the three vowels (I , E ,O) as one event. This one event needs to be arranged along with other five letters(D, R, C, T, R) in 6!/2!(6×5×4×3×2×1/2)=360 ways.
Hence total number of words that can be formed=6 ×360= 2160.
The three vowels(I, E , O) can be arranged among themselves in 3! (3×2×1)=6 ways.
Next, we can consider the three vowels (I , E ,O) as one event. This one event needs to be arranged along with other five letters(D, R, C, T, R) in 6!/2!(6×5×4×3×2×1/2)=360 ways.
Hence total number of words that can be formed=6 ×360= 2160.
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