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How many six-digit numbers can be made by using each of the digits 1–6 exactly once? (b) How many of those numbers are smaller than 300 000? [5 marks]


A baby has nine different toy animals. Five of them are red and four of them are blue. She arranges them in a line so that the colours are arranged symmetrically. How many different arrangements are possible? [7 marks]


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First question is relatively easy. It is permutations of 6 numbers, which is 6!. A third of these numbers start with 1 or 2, so 6!/3 = 240 of them are less than 300000.

For a line with an odd number of animals, it is symmetric only when the 5th (middle) one is red. Then to the left and right, there are two red and two blue on each side. There are 4C2 = 6 ways to arrange two red and two blue (namely RRBB RBRB RBBR, BBRR, BRBR, BRRB). A key conceptual understanding to make is that for every arrangement of the colors, the specific animals can be swapped around and that would create a new line. Among the 5 red, they can be swapped 5! ways. Among the 4 blue, they can be swapped 4! ways. The answer is 6(5!)(4!) = 17280. 

Remember that in the HL Core, counting problems are rare on the actual exam so if you find them conceptually difficult it is not the end of the world.  

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