Permutation and Combination

No. of ways of selecting r objects out of n objects = nCr
No. of ways of arranging n objects = n!
No. of ways of arranging n objects with a identical & b identical = n!
a! b!
No. of ways of arranging r objects out of n objects
= n(n–1)(n–2) ... (n – (r – 1)) = nPr = nCr r!
No. of ways of arranging n objects in a circle = (n – 1)!

Sample questions:
1. Choosing People:
Que: A team of 4 is to be chosen from a group consisting of Anne and Bob and 4 other people. In how
many ways can this be done if
(i) there are no restrictions?
(ii) Anne must be in the team?
(iii) Anne and Bob must both be in the team?
(iv) at most one of Anne and Bob in the team?
(v) Anne or Bob or both are in the team?
Ans:
(i) No. of ways of choosing 4 people = 6C4 = 15
(ii) No. of ways of choosing the other 3 people = 5C3 = 10
(iii) No. of ways of choosing the other 2 people = 4C2 = 6
(iv) Total no. of ways – no. of ways with Anne & Bob both in the team = 15 – 6 = 9
(v) No. of ways with Anne in the team + no. of ways with Bob in the team – no. of ways with
both in the team = 10 + 10 – 6 = 14

2. Choosing from Different Types of People (e.g. Boys & Girls)
Que: A team of 3 is to be chosen from a group of 3 boys and 4 girls. How many ways can this be done if
(i) there are no restrictions?
(ii) there must be exactly 1 boy?
(iii) there must be at least 1 boy?
(iv) there must be at least 1 boy and at least 1 girl?

Ans: (i) No. of ways of choosing 4 people = 7C3 = 35
(ii) No. of ways of choosing 1 boy and 2 girls = 3C1 4C2 = 18
(iii) No. of ways = Total no. of ways – no. of ways with no boys = 35 – 4C3 = 35 – 4 = 31
Note: It is wrong to say no. of ways = 3C1 6C2 = 45
(iv) No. of ways = Total no. of ways – no. of ways with no boys – no. of ways with no girls
= 35 – 4C3 – 3C3 = 35 – 4 – 1 = 30
Note: It is wrong to say no. of ways = 3C1 4C1 5C1 = 60