By D., V.T. Sos, T. Szonyi eds. Miklos
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Extra resources for Combinatorics, Paul Erdos is Eighty Volume 1
We have 10 digits. We have to make all five digit numbers. The total such numbers is equal to 105. Here we have to make slips for these many numbers. The numbers made of digits 0, 1, 6, 8 and 9 can be read upside down or right side up. And, there are 55 many such five-digit numbers (all those five-digit numbers made of digits 0, 1, 6, 8 and 9). Out of these 55 many numbers, however, there are some numbers that read the same either upside down or right side up. For example, 91816, and there are 3 × 52 such numbers (center place filled with 0, 1 or 8).
The generating function for x is (1 + x + x2 + x3 + x4 + x5), for y is (1 + y + y2 + y3 + y4) and for z is (1 + z + z2 + z3). If we replace y and z by x, we get the generating function for the above problem as f(x) = (1 + x + x2 + x3 + x4 + x5)(1 + x + x2 + x3 + x4)(1 + x + x2 + x3) The coefficient of x6 in f(x) is the required number, and this number is 18. In the second part, at least one book of each colour has to be selected, so generating function f(x) is given as f(x) = (x + x2 + x3 + x4 + x5)(x + x2 + x3 + x4)(x + x2 + x3) The coefficient of x6 in f(x) is the required number, and this number is 9.
Possible lines. Similarly, there are 4 ! lines in which the boy, dog, and man appear in this order. So, by the addition rule, there are 4 ! + 4 ! = 48 lines in which the dog (and only the dog) is between the man and the boy. 50. In how many ways can ten adults and five children stand in a circle so that no two children are next to each other ? Solution. Arrange the adults into a circle in one of 9 ! ways. There are then 10 locations for the first child, 9 for the second, 8 for the third, 7 for the fourth, and 6 for the fifth.