Find the total number of ways in which six ‘+’ and four ‘-‘ signs can be arranged in a line such that no two ‘-‘ signs occur together.
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.