Total no. of ways in which six plus (+) and four minus (-) signs can be arranged so that no two minus signs come together is ?? The options are 20,35,42,56...

My attempt:

I fixed position of plus signs as follows leaving a space in between as follows:

(plus) (plus) (plus) (plus) (plus) (plus)

This can be done in only way since all the plus signs in identical...now i have to fill seven places left with four minus signs which can can be done how many ways?? Heres where the problem comes in...is the no.of ways of performing this task equal to (7 C 4) or 7!/4! ???

I think it should be the former....but am i correct or wrong....plz suggest.....Thanks in advance!!


Firstly, let's place all the plus signs in a line, with a gap between each pair of plus signs (2 gaps at the ends too) _ + _ + _ + _ + _ + _ + _

Now our goal is to place the 4 minus signs in any 4 out of those 7 blanks In this way, no 2 minus signs will end up together

Hence, the number of ways of selecting 4 gaps out of 7 is ( 7 C 4 ) = 35

  • $\begingroup$ Ok thnx...it cleared my doubt $\endgroup$ – SirXYZ Oct 10 '16 at 17:53

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