By C. Vasudev

**Read Online or Download Combinatorics and Graph Theory: As Per U.P.T.U. Syllabus PDF**

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**Extra resources for Combinatorics and Graph Theory: As Per U.P.T.U. Syllabus**

**Sample text**

1 We read n ! as “n factorial”. It is true that 4 ! = 24 and 6 ! = 720 but frequently we leave our answers in factorial form rather than evaluating the factorials. Nevertheless, the relation n ! ] enables us to compute the values of n ! for small n fairly quickly. For example : 0 ! = 1, 1 ! =2 3 ! = 6, 4 ! = 24, 5 ! = 120 6 ! = 720, 7 ! = 5040, 8 ! = 40320 9 ! = 362880, 10 ! = 3628800, 11 ! = 39916800. 10. Enumerating r-permutations without repetitions P(n, r) = n(n – 1) ...... (n – r + 1) = n!

92. Find the total number of selections of at least one red ball from 4 red balls and 3 green balls, if (a) the balls of the same colour are different (b) the balls of the same colour are identical. Solution. (a) From 4 different red balls and 3 different green balls, we have to find number of selections taking at least one red ball and any number of (including 0) 3 green balls. The total number of ways of selecting at least one red ball from 4 different red balls = 4C1 + 4C2 + 4C3 + 4C4 = 15. Corresponding to each of these selections, the number of ways of selecting green balls = 3C0 + 3C1 + 3C2 + 3C3 = 8.

A) Since the bits 0 or 1 can repeat, the eight positions can be filled up either by 0 or 1 in 28 ways. Hence the number of bytes that can be formed is 256. (b) Keeping two positions at the beginning by 11 and the two positions the end by 11, there are four open positions which can be filled up in 24 = 16 ways. Hence the required number is 16. (c) Keeping two positions at the beginning by 11, the remaining six open positions can be filled up by 26 = 46 ways. Hence the required number is 64 – 16 = 48.