Seating Arrangement I

Probability Level 2

A small class of nine boys are to change their seating arrangement by drawing their new seat numbers from a box. After the seat change, what is the probability that there is only one pair of boys who have switched seats with each other and only three boys who have unchanged seats?

Source: Philippines Mathematical Olympiad

1 14 \frac{1}{14} 1 21 \frac{1}{21} 1 48 \frac{1}{48} 1 24 \frac{1}{24}

Ir J by IR J , 21, Philippines

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1 solution

Chew-Seong Cheong
Nov 18, 2018
  • The total number of new seating arrangements N T = 9 ! N_T = 9!
  • The number of ways to choose the 3 boys with unchanged seats N a = ( 9 3 ) N_a = \dbinom 93
  • The number of ways the pair of boys who switch seats from the remaining 6 boys N b = ( 6 2 ) ( 4 2 ) ( 2 2 ) = ( 6 2 ) ( 4 2 ) N_b = \dbinom 62 \dbinom 42 \dbinom 22 = \dbinom 62 \dbinom 42

Then the required probability P = N a N b N T = ( 9 3 ) ( 6 2 ) ( 4 2 ) 9 ! = 1 48 P = \dfrac {N_aN_b}{N_T} = \dfrac {\binom 93 \binom 62 \binom 42}{9!} = \boxed{\dfrac 1{48}} .

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@IR J , I have edited your problem wording as above. I believe PMO is Philippines Mathematical Olympiad. Is it right?

Chew-Seong Cheong - 2 years, 6 months ago

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Yes sir, it's right. Thanks!

IR J - 2 years, 6 months ago

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