Problem 1 :
Find the total number of subsets of a set with
[Hint: ^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + · · · + ^{n}C_{n} = 2^{n}]
(i) 4 elements (ii) 5 elements (iii) n elements
Solution :
(i) 4 elements
^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + · · · + ^{n}C_{n} = 2^{n}
here n = 4
= ^{ 4}C_{0} + ^{4}C_{1} + ^{4}C_{2} + ^{4}C_{3 }+ ^{4}C_{4}
= 2^{4}
= 16
(ii) 5 elements
^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + · · · + ^{n}C_{n} = 2^{n}
here n = 5
= 2^{5}
= 32
(iii) n elements
^{n}C_{0} + ^{n}C_{1} + ^{n}C_{2} + · · · + ^{n}C_{n} = 2^{n}
here n = n
= 2^{n }elements
Problem 2 :
A trust has 25 members.
(i) How many ways 3 officers can be selected?
(ii) In how many ways can a President, Vice President and a Secretary be selected?
Solution :
(i) Out of 25 members only 3 officers can be selected.
number of ways of selecting 3 officers = ^{25}C_{3}
= 25!/22! 3!
= (25 ⋅ 24 ⋅ 23)/6
= 2300
(ii) To select a president, we have 9 options
to select vice president, we have 8 options
to select secretary, we have 7 options
total number of ways = 9 ⋅ 8 ⋅ 7 = 504
Hence the answer is 504.
Problem 3 :
How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?
Solution :
Out of 10 members, we have to select a chair person. So we have 10 options to select a chair person. 9 options to select a secretary.
After selecting a chair person and secretary, we have 8 member. Out of 8, we have to select 4 persons.
Hence the answer is (10 ⋅ 9) ^{8}C_{4}
Problem 4 :
How many different selections of 5 books can be made from 12 different books if,
(i) Two particular books are always selected?
(ii) Two particular books are never selected?
Solution :
(i) Since two particular books are always selected, we may select remaining 3 books out of 10 books.
^{10}C_{3 }= 10!/(7! 3!) = (10 ⋅ 9 ⋅ 8)/( 3 ⋅ 2)
= 120
Hence the answer is 120.
(ii) Since two particular books are never selected, we may select 5 books out of 10 books.
^{10}C_{5} _{ }= 10!/(5! 5!) = (10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6)/(5 ⋅ 4 ⋅ 3 ⋅ 2)
= 252
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