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Exercise 6.3 · Q7

Q.A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that

(i) all the sisters sit together
(ii) no two sisters sit together
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Use the block method for "together" and the gap method for "no two together".

Block/grouping method: tie the group that must stay together into one super-item; arrange super-items, then arrange members within the block internally. Gap method: arrange the unrestricted group first, then insert the restricted group into the gaps created (ends included) so no two of them are adjacent. nPr=n!(n−r)!^nP_r=\dfrac{n!}{(n-r)!}.

(i) All 4 sisters sit together

  1. Treat the 4 sisters as a single block. Now we are arranging 66 brothers + 1+\ 1 block =7=7 entities in a row: 7!7! ways.
  2. Within the block, the 4 sisters can be permuted among themselves: 4!4! ways.
  3. Total =7!×4!=5040×24=7!\times4! = 5040\times24.
  4. Compute: 5040×24=1209605040\times24=120960.

(ii) No two sisters sit together

  1. First arrange the 6 brothers in a row: 6!=7206!=720 ways. This creates 77 gaps (one before, one after, and one between each pair of brothers). …

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