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Exercise 6.4 · Q6

Q.A box contains 6 red and 7 white balls. Determine the number of ways in which 4 red and 3 white balls can be selected.

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The red-ball selection and the white-ball selection are independent choices, so multiply the two combination counts (multiplication principle).

[!FORMULA] nCr=n!r!(n−r)!^{n}C_{r}=\dfrac{n!}{r!(n-r)!} gives the number of ways to choose rr objects from nn available objects, order irrelevant.

  1. Ways to choose 4 red balls from 6 red balls =6C4=6!4! 2!=6×52×1=15={}^{6}C_{4}=\dfrac{6!}{4!\,2!}=\dfrac{6\times5}{2\times1}=15.
  2. Ways to choose 3 white balls from 7 white balls =7C3=7!3! 4!=7×6×53×2×1=35={}^{7}C_{3}=\dfrac{7!}{3!\,4!}=\dfrac{7\times6\times5}{3\times2\times1}=35. …

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