For the 3rd questions (Q.9 b)) there are 3 sheep remaining and 4 pens remaining. Shouldn't we use permutation 4P3 here rather than 4 factorial? If my observation is wrong then could you please explain why you put 4! there instead?
Hi Lara, Thanks for your question. For Question 9b, we are using the Complimentary Method. So we take the No Restrictions Case of 6! and subtract from it the case where Sheep A & B share a boundary. To answer your question directly, 4! and 4P3 are essentially the same in value, 24. Both 4! and 4P3 in the IB working will work. For 4!, there are 4 pens remaining for 4 sheep, so we do 4P4 or 4! = 24. If you want to use 4P3=24, that is fine too. It means arranging 3 sheep in 4 remaining slots, leaving no choice for the remaining empty pen. Either way, the answer is the same. Hope this helps!
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Thanks so much
For the 3rd questions (Q.9 b)) there are 3 sheep remaining and 4 pens remaining. Shouldn't we use permutation 4P3 here rather than 4 factorial? If my observation is wrong then could you please explain why you put 4! there instead?
Hi Lara,
Thanks for your question. For Question 9b, we are using the Complimentary Method. So we take the No Restrictions Case of 6! and subtract from it the case where Sheep A & B share a boundary.
To answer your question directly, 4! and 4P3 are essentially the same in value, 24. Both 4! and 4P3 in the IB working will work. For 4!, there are 4 pens remaining for 4 sheep, so we do 4P4 or 4! = 24. If you want to use 4P3=24, that is fine too. It means arranging 3 sheep in 4 remaining slots, leaving no choice for the remaining empty pen.
Either way, the answer is the same. Hope this helps!