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
A.
$$\frac{{7!}}{{3!}}$$
B.
$$6!\, \times \frac{{7!}}{{3!}}$$
C.
$$35$$
D.
None of these
Answer :
$$35$$
Solution :
$$' - '$$ signs will be put between two $$' + '$$ signs or at the two ends.
There are $$7$$ places for four $$' - '$$ signs. So, the required number of ways
$$ = {\,^7}{C_4} = \frac{{7!}}{{4!\,3!}}$$
(there being no arrangement as the $$' + '$$ signs are identical as well as $$' - '$$ signs are identical).
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
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