Permutation & Combination Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

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.

Answer: (d)

CORPORATION= 11 letters

'O' comes thrice, 'R' twice.

∴ total no. of ways = ${11!}/{3!2!}$ = 3326400

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.

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

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