(a). The number of Distinct letters in the word "wombat " = 6
The number of different ways can the letters of the word "wombat" be arranged
=P(6,6)
=(6−6)!6!=0!6!=16×5×4×3×2×1=720
(b) The letter "w and o" can treat as one Unit
So, the number letters = 4 letters + one unit of w and 0= 5 things
The number of ways 5 things can be arranged is
=P(5,5)=(5−5)!5!=1120=120
(c) The number of different 3-letter words can be formed is
P(6,3)=(6−3)!6!=3!6×5×4×3×2×1
=3×2×16×5×4×3×2×1=6×5×4=120
If w must be the first letter of any such 3-letter word, we need to choose 2 letters from the remaining 5 letters.
Number of different ways
=P(5,2)=(5−2)!5!=3!5×4×3×2×1
=3×2×15×4×3×2×1=5×4=20
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