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) = 7P4


But all 4 ‘-’ signs are identicals hence a number of ways of arrangements of 4 ‘-’ signs among 7 positions = 7P4 / 4!


A total number of arrangments of + and – signs are given by 7P4 / 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 7C4 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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