Counting, Permutations & Combinations

COUNTING , PERMUTATIONS & COMBINATIONS

5.PERMUTATIONS & COMBINATIONS OF OBJECTS WHICH ARE NOT ALL DIFFERENT

In this subtopic, when finding the total number of combinations, different cases have to be considered if the word has repeated letters (the number of repeated letters is very important). The total number of permutations depends on the cases formed in the combinations. This topic is best understood using a variety of examples. E.gs.

(1)How many (a)combinations (b) arrangements are there of 3 letters chosen at random from the word BIOLOGY.

(a) 3 cases have to be considered.

with no O's : 5C3 = 10 (BILGY, 3/5)

with 1 O : 5C2 = 10 (O & 2/5 from BILGY)

with 2 O's : 5C1 = 5 (OO & 1/5 from BILGY)

No. of combinations = 10 + 10 + 5 + 25

(b) Case 1 & Case 2 have 3 distinct letters while case 3 has a repeated O to cater for (divide by 2!)

No. of arrangements = 10 x 3! + 10 x 3! + 5 x 3!/2!

= 135.

(2) How many (a) combinations (b) arrangements ,are there of 4 letters chosen at random from the word SAUCEPAN?

(a) 3 cases have to be considered.

with 2 A's : 6C2 = 15 (AA & 2/6 from SUCEPN)

with 1 A : 6C3 = 20 (A & 3/6 from SUCEPN)

with no A's : 6C4 = 15 (4/6 from SUCEPN)

No. of combinations = 15 + 20 + 15 = 50

(b) Case 2 & case 3 have 4 distinct letters while case 1 has repeated

No. of arrangements = 15 x 4!/2! + 20 x 4! + 15 x 4!

= 1020

(3) How many (a) combinations (b) permutations are there of 3 letters chosen at random from the word COFFEE.

(a) 3 cases have to be considered

with 2F : 3C1 = 3 (FF & 1/3 from COE)

with 2 E : 3C1 = 3 (EE & 1/3 from COF)

3 distinct : 4C 3 = 4 (3/4 from COFE)

No. of combinations = 3 + 3 + 4 = 10

(b)Case 1 & 2 have repeated letters while case 3 has 3 distinct

no. of arrangements = 3 x 3!/2! + 3 x 3!/2! + 4 x 3!

= 42