Answer :

METHOD: 1 (Method of Permutations)

Given 6 ‘+’ and 4 ‘-’ signs to arrange.

To find: Number of arrangements such that no two ‘-’ comes together.

A number of ways of arrangements of 6 ‘+’ signs = 1 (Since all ‘+’ signs are identical).

Arranging all 6 + signs in such a way that now minus sign will only occupy a place in between or edge of + signs.

For example- “P+P+P+P+P+P+P” Arranging 6 + signs gives 7 positions in which a ‘-’ sign can be placed (represented here as P).

Now, number of ways of arrangements of 4 ‘-’ signs in 7 places (Placed all +’s in an alternate manner leading 7 positions) = ^{7}P_{4}

But all 4 ‘-’ signs are identicals hence a number of ways of arrangements of 4 ‘-’ signs among 7 positions = ^{7}P_{4} / 4!

A total number of arrangments of + and – signs are given by ^{7}P_{4} / 4!

= 5 x 7

= 35

Hence, a total number of arrangements of + and – signs is 35.

METHOD: 2 (Method of combinations)

All 6 + signs can be arranged in 1 way (All are identical).

Now we have 7 places and 4 ‘-’ signs: So we have to select a position and arrange 4 signs among 7 places. In how many ways can we did it? We can do it in ^{7}C_{4} number of ways which indeed equal to

= 5 x 7

= 35

Hence, total number of arrangements of + and – signs is 35.

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