By C. Vasudev
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Additional info for Combinatorics and Graph Theory (as per U.P.T.U. Syllabus)
Let us draw a chart illustrating the possible male-female and junior senior constitution of the committee. Juniors Female Seniors Male Female Male Number of ways of selecting 0 1 3 2 5 4 2 3 C(30, 0) C(35, 3) C(25, 5) C(20, 2) C(30, 1) C(35, 2) C(25, 4) C(20, 3) 2 1 3 4 C(30, 2) C(35, 1) C(25, 3) C(20, 4) 3 0 2 5 C(30, 3) C(35, 0) C(25, 2) C(20, 5) Thus, the total number of ways is the sum of the terms in the last column: C(30, 0) C(35, 3) C(25, 5) C(20, 2) + C(30, 1) C(35, 2) C(25, 4) C(20, 3) + C(30, 2) C(35, 1) C(25, 3) C(20, 4) + C(30, 3) C(35, 0) C(25, 2) C(20, 5).
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. The answer is 9 ! 6) = 9 ! P(10, 5). 51. Find (a) P(5, 3), P(4, 4) and P(7, 2) (b) 20 !
In how many ways can ten adults and five children stand in a line so that no two children are next to each other ? Solution. , j, XD XJ XH XC XI XE XB XA XG XF X, then X’s representing the 11 possible locations for the children. For each such line, the first child can be positioned in any of the 11 spots, the second child in any of the remaining 10, and so on. 7 = P(11, 5) ways. For each such positioning, there are 10 ! , J, so, by the multiplication rule, the number of lines of adults and children is 10 !
Combinatorics and Graph Theory (as per U.P.T.U. Syllabus) by C. Vasudev