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Exercise 6.4 · Q10

Q.In how many ways can 7 plus (+) signs and 5 minus (–) signs be arranged in a row so that no two (–) signs are together.

Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
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Fix the 7 identical plus signs first to create gaps between and around them, then choose gaps to place the minus signs so no two minus signs end up adjacent.

[!FORMULA] For pp identical objects of one kind creating p+1p+1 gaps, and mm identical objects of a second kind to be placed with no two together, the number of ways =p+1Cm={}^{p+1}C_{m} (choose mm of the p+1p+1 gaps, at most one object per gap since signs of the same kind are identical).

  1. Arrange the 7 identical (+)(+) signs in a row: since they are identical, there is only 11 way to arrange them, and they create 7+1=87+1=8 gaps: one before the first ++, one between each consecutive pair of ++ signs (6 such gaps), and one after the last ++ — total 6+2=86+2=8 gaps.
  2. To ensure no two (−)(-) signs are together, at most one (−)(-) sign can go into each gap.
  3. We must place 55 identical (−)(-) signs into 55 of these 88 gaps (choosing which gaps, order irrelevant since the signs are identical): 8C5^{8}C_{5} ways. …

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