Page 177 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes
Example: Now presume that we have to create a team of 11 players out of 20 players, This is
an instance of combination, since the order of players in the team will not consequence in a
transformation in the team. Regardless of in which order we list out the players the team will
continue to be the same! For a different team to be created at least one player will have to be tainted.

Notes The dissimilarity among combinations and permutations is in combinations you
are counting groups and in permutations you are counting different methods to assemble
items with respect to order.
The n and the r signify the similar thing in both the permutation and combinations, but the
formula varies.

!
Caution The combination has an additional r! in its denominator.

Example: C1—r = n: How many ways can a committee of 7 people be drawn from a
group of 7 people?

C(7, 7) = 7!/[(7 – 7)!’”7!] = 7!/[0!’”7!] = 7!/7! = 1.
In other words, there is exactly 1 way to choose all 7 people to be on the committee since order
doesn’t matter.

Example: C2—r < n: How many ways can 2 co-chairpersons be chosen from a committee
of 7 people?
C(7, 2) = 7!/[(7 – 2)!×2!] = 7!/[5!×2!] = 7×6×5×4×3×2×1/(5×4×3×2×1 )(2×1) = 7×6/2×1 = 7×3 = 21.
In other words, there are 21 ways to select a pair of co-chairs from a committee of 7 people.
Relating the above scenario to permutations, note that there are 7 ways to choose the first co-
chair and 6 ways to choose the second co-chair. So if order mattered, we would have 42 possible
choices. However, the combination of 2 co-chairs is the same, no matter which person was
selected first. Thus, we must divide by the by the 2! ways the two chair persons could have been
selected. Applying the actual formula, we have
P(7, 2) = 7!/(7 – 2)! = 7!/5! = 7×6 = 42.
But
C(7, 2) = 7!/[(7 – 2)!×2!] = 7!/[5!×2!] = (7×6)/(2×1) = 42/2 = 21.

Example: C3: In a club with 8 males and 11 female members, how many 5-member
committees can be chosen that have 4 females?
Solution: Since order doesn’t matter, we will be using our combinations formula.
How many ways can the 4 females be chosen from the 11 females in the club?
C(11, 4) = 11!/[(11 – 4)! ×4!] = 11!/(7! ×4!) = (11×10×9×8)/ (4×3×2×1) = 330 ways.
But this gives us only the 4 female members for our 5-member committee. For a 5-member
committee with 4 females, how many males must be chosen? Only 1. How many ways can the
1 male be chosen from the 8 males in the club?
C(8, 1) = 8!/[(8 – 1)! × 1!] = 8!/(7! × 1!) = 8 ways.

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