permutation problems plz solve
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Hi Abhijeet ans to 4) let us consider ABCD to be in order. now we have to insert 2 people between them. 1st case: the other two persons(E&F) will be together _ _ A _ _ B _ _ C _ _ D _ _ no.of ways 5C1*2!=10 ways 2nd case : the other two persons will not be together _ A _ B _ C _ D _ no. of ways 5C2* 2!=20 ways total ways=10+20=30 the mistake made by kvenkan was that if we insert the other people in position 1 & 5 or 2 & 5(see fig below) there is no difference in the arrangement. 12A34B56C78D910 Hope you understood |
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