Permutation & Combination Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts
PERMUTATIONS & COMBINATION PRACTICE TEST [1 - EXERCISES]
Permutation & Combination Model Questions Set 1
Question : 21
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
a) 196
b) 280
c) 346
d) 140
e) None of these
Answer »Answer: (a)
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.
∴ No. of choices available to the student
= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.
Question : 22
In how many different ways can the letters of the word CORPORATION be arranged?
a) 831600
b) 1663200
c) 415800
d) 3326400
e) 207900
Answer »Answer: (d)
CORPORATION= 11 letters
'O' comes thrice, 'R' twice.
∴ total no. of ways = ${11!}/{3!2!}$ = 3326400
Question : 23
In how many ways can 7 persons be seated at a round table if 2 particular persons must not sit next to each other ?
a) 480
b) 240
c) 720
d) 5040
e) None of these
Answer »Answer: (a)
Total no. of unrestricted arrangements = (7 – 1) ! = 6 !
When two particular person always sit together, the total no. of arrangements = 6! – 2 × 5!
Required no. of arrangements = 6! – 2 × 5!
= 5! (6 – 2) = 5 × 4 × 3 × 2 × 4 = 480.
Question : 24
the committee should include all the 3 Readers?
a) 21
b) 180
c) 55
d) 90
e) None of these
Answer »Answer: (c)
Total number of ways to form committee
= $^5C_2 × ^6C_0 × ^3C_3 + ^5C_1 × ^6C_1 × ^3C_3 + ^5C_0 × ^6C_2 × ^3C_3$ = 10 + 30 + 15 = 55
Question : 25
There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is
a) 40
b) 36
c) 48
d) 24
e) None of these
Answer »Answer: (c)
Let the four candidates gets the votes x, y, z and w such that
x + y + z + w = 51 ...(i)
Here x ≥ 0 , y ≥ 0, z ≥ 0, w ≥ 0
The number of solutions of the above equation in this case is same as the number of ways in which the votes can be given if at least no two candidates get equal number of votes.
(Note : The number of ways in which n identical things can be distributed into r different groups = $^{n + r – 1}{C_{r - 1}}$)
∴ Total number of solutions of eqn. (i)
= $^{5 + 4 – 1}C_{4 – 1}$ = $^8C_3$ = 56
But in 8 ways the two candidate gets equal votes which are shown below :
(2, 2, 1, 0), (2, 2, 0, 1), (0, 2, 2, 1), (1, 2, 2, 0), (0, 1, 2, 2), (1, 0, 2, 2), (2, 0, 1, 2), (2, 1, 0, 2)
Hence the required number of ways = 56 – 8 = 48
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Permutation & Combination Model Questions Set 1 Online Quiz
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