1) there are 2 R ; 4 E and 1 F
number of ways to arrange =7!/(2!*4!*1!)=105
2)
number of ways 2 R's are together =6!/(1!*4!*1!) =30
hence number of ways two R's separated=105-30=75
3)
number of ways 4' E's are separated =3!/(2!*1!)=3
hence contains subword EE =105-3=102
4)
number of ways it begins with letter R =6!/(1!*4!*1!) =30
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