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Permutation & combination

Questions



Q1. In how many ways can a person invite one or more of his five friends to a party ?

(a) 31

(b) 32

(c) 30

(d) 33

Solution. a
= 2n - 1 = 25 - 1 = 31

Note : nC1 + nC2 + nC3 + -----------+ nCn = 2n - 1
Solution

Q2. In how many ways 52 playing cards be distributed to four players giving 13 cards each ?

(a) 52!

(b)
52!/(13!)4


(c) 13!

(d) (13!)4

Solution.b
Total number of ways =
52!/(13!)4


Note : The number of ways in which 4P things can be divided equally into four distinct groups is =
(4P!)/(P!)4
Solution

Q3. In a foundation examination , a candidate is required to pass four different subjects . In how many ways
can he fail ?

(a) 15

(b) 16

(c) 17

(d) 18

Solution. a
2n - 1 = 24 - 1 = 15

Note : nC1 = Chance to fail in one subject .

nC2 = Chance to fail in two subject .

nC3 = Chance to fail in three subject .

nC4 = Chance to fail in four subject .

nC1 + nC2 + nC3 + nC4 = 24 - 1
Solution

Q4. 5C3 + 2(5C2) + 5C3 = x , find x .

(a) 10

(b) 40

(c) 24

(d) 30

Solution. b
= 5C3 + 5C2 + 5C2 + 5C3

= 6C3 + 6C3

= 2 (6C3) = 40
Solution

Q5. The maximum number of points of inter-section of 10 circles will be .

(a) 40

(b) 900

(c) 90

(d) 60

Solution.c
Formula = nC2 x 2

= 10C2 x 2

= 90
Solution