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Worked Examples · Example 32

Q.In how many ways can 4 men and 4 women be seated at a round table if

(i) no restriction is imposed
(ii) two particular women must sit together
(iii) all women must sit together
Yanam CbseNCERTSubjective· 3mImportance★★★★★est
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✓ Free question

Round table with 4 men + 4 women: unrestricted =5040=5040; two particular women together =1440=1440; all women together =576=576.

Circular arrangement of nn distinct people (rotations identical) =(n−1)!=(n-1)!. For a "must sit together" restriction, glue the restricted people into one block, arrange the resulting units circularly using (units−1)!(\text{units}-1)!, then multiply by the internal arrangements of the block.

  1. Total people =4=4 men +4+4 women =8=8, all distinct.
  2. (i) No restriction: circular arrangement of all 8:

(8−1)!=7!=5040(8-1)!=7!=5040

  1. (ii) Two particular women sit together: glue those 2 women into 1 block. Units to arrange =4=4 men +2+2 remaining women +1+1 block =7=7 units. Circular arrangement of 7 units: (7−1)!=6!=720(7-1)!=6!=720. Internal order of the 2 glued women: 2!=22!=2. Total =720×2=1440=720\times2=1440.
  2. (iii) All 4 women sit together: glue all 4 women into 1 block. Units to arrange =4=4 men +1+1 block =5=5 units. Circular arrangement of 5 units: (5−1)!=4!=24(5-1)!=4!=24. Internal order of the 4 women inside the block: 4!=244!=24. Total =24×24=576=24\times24=576.
  3. Self-check: 576<1440<5040576<1440<5040 — as expected, each added restriction shrinks the count. ✓
✓Final answer

(i) 50405040 ways;

(ii) 14401440 ways (two particular women together);

(iii) 576576 ways (all women together).

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