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Question

The total number of ways in which six $$'+'$$ and four $$'-'$$ signs can be arranged in a line such that no two $$'-'$$ signs occur together is ______.


Solution

$$'+'$$ signs can be put in a row in one way creating seven gaps shown as arrows:
Now $$4 '-'$$ signs must be kept in these gaps. So, no two $$'-'$$ signs should be together.
Out of these $$7$$ gaps $$4$$ can be chosen in $$^{7}C_{4}$$ ways. Hence, required number of arrangements is
$$^{7}C_{4} = ^{7}C_{3} = \dfrac {7\times 6\times 5}{3\times 2\times 1} = 35$$.
1569055_145674_ans_46501f63d1314d05a1a499c789368078.jpg

Mathematics

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