1.4 - Number the twelve edges of a cube with numbers 1 through 12 in such a way
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1.4 - Number the twelve edges of a cube with numbers 1 through 12 in such a way that the sum of the three edges meeting at each vertex is the same for each vertex.
1.2 Suppose we are given n disjoint sets Si. Let the first have ai elemen ond a2, and so on. Show that the number of sets that contain at most one from each S, is equal to (a1 + 1) a). ts,t sec +1). Apply this result to the following number theoretic problem: Let n = r" p be the prime decomposition of n. Then n has exactly t(n) II(ai 1) divisors. Conclude that n is a square if and only if t(n) is odd. Let N = { 1, 2, . . . , 100, and let A be a subset of N with IA-55. Show that A contains two numbers a and b such that a -b 9. Does this hold as well for |A] = 54? 1.4 Number the twelve edges of a cube with the numbers 1 through 12 in such a way that the sum of the three edges meeting at each vertex is the same for each D> 1.3 vertex. D 1.5 In the parliament of country X there are 151 seats and three political parties. How many ways (i, j, k) are there of dividing up the seats such that no party has an absolute majority? 1.6 How many different words can be made from permutations of the letters in ABRACADABRA? 1.7 Show that 1!+2!+ + n! for n > 3 is never a square. 1.8 Show that for the binomial coefficients (H) the following holds:Explanation / Answer
It is impossible to number the twelve edges of a cube with numbers 1 through 12 in such a way that the sum of the three edges meeting at each vertex is the same for each vertex because let suppose S be the sum of the three edges meeting at each vertex. The total of all such eight vertices is 8S as sum is same on each. But this counts each edge twice if we draw the figure and make some effort to count it, and the sum of the 12 edges is 1+2++12=78, so 8S=2*78 or S=19.5. But S is the sum of any three vertices ranging from 1 to 12 so S can't be 19.5. Which is a contradiction.
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