Test:- Permutations And Combinations - 11 - Mathematics MCQ

Total number of subsets
= The number of selections of at least two elements inluding m, n and natural numbers lying between m and n.
= total number of selections fro n - m - 1 different things
= 2^n-m-1

Total number of ways = m x m x ... to n times = mⁿ.

Total number of ways = 6 x 6 x ... to n times = 6ⁿ
Total number of ways to show only even numbers
= 3 x 3 x ... to n times = 3ⁿ.
∴ the required number of ways = 6n - 3n.

0 cannot go in the units place.
∴ the required number o f numbers = 2 * 3⁵.

The matrix will be of the order 4 M or 1 x 4 or 2 x 2.
For each order, the number of different coordinates ⁿP₃.
∴ the required number of triplets = n³-n(n-1)(n-2)

Total - none equal = n³-ⁿP₃

From every arrangement of 7 letters we get a pair by putting a sign after the four letters from the left.
∴ the required number of pairs = 7!/2!2!

Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repreated.
The number of selections of the sixth digit = ⁵C₁= 5.
∴ the required number ot numbers = 

∴ the required number of divisors
= the number of selections of one 2 from four 2’s, any number of 3’s from one 3 and any number of 5’s from one 5.
= 1 x 2 x 2 = 4.

The number of marches in the first round = ⁶C₂ + ⁶C₂
The number of matches in the next round = ⁶C₂
The number of matches in the semifinal round = ⁴C₂.
∴the required number of matches

Note: For “best of three” at least two matches are played.

The number of selection of 4 persons including A, B = ⁶C₂.
Considering these four as a group, number of ar rangements with the other four = 5!
But in each group the number of arrangements = 2! x 2!
∴ the required number of ways =

The number of committees of 4 gentlemen =⁴C₄ = 1
The number of committees of 3 gentlemen, 1 wife = ⁴C₃ x ¹C₁
(∴ after selecting 3 gentlemen only 1 wife is left who can be included).
The number- of committees of 2 gentlemen, 2 wives
= ⁴C₂ x ²C₂
The number of committees of 1 gentlemen. 3 wives
= ⁴C₁ x ³C₃.
The number of committees of 4 wives = 1
∴ the required number of committes = 1 + 4 + 6 + 4+1 = 16

The committee will consist of 4 gentlemen and 1 lady or 3 gentlemen and 2 ladies.
∴ the number of committes = 

If r questions are solved by each student then the number of possible selections of questions is ²⁰C_r
∴ the number of students =²⁰C_r
(∴ each student has solved different combination of questions)
∴ the maximum number of students = maximum value of ²⁰C_r = ²⁰C₁₀, because ²⁰C₁₀ is the largest among ²⁰C₀, ²⁰C₁, ....²⁰C₂₀ - being the middle one. 

Each place can be filled in 3 ways.
so, n places can be filled in 3 x 3 x ... to n times = 3ⁿ.
3⁶= 729, 3⁷=2187. So, for 3^n-3 900 the smallest integral n is 7.

24 = 2 x 3 x 4 , 2 x 2 x 6 , 1 x 6 x 4, 1 x 3 x 8 , 1 x 2 x 12, 1 x 1 x 24
(as product of three positive integers)
∴ the total number of positive integral solutions of xyy. = 24 is equal to 

Any two of the factors in each factorization may be negative.
∴ the number of ways to associate negative sign in each case is ³C₂, i.e., 3.
∴ the total number of integral solutions = 30 +3 x 3 = 120 

The number of ways to fill the three even places by 4 constonants = ⁴P₃.
After filling the even places, remaining places can be filled in ⁴P₄ ways.
So, the required number of words = ⁴P₃ x ⁴P₄.
 

pivot has been created by chairman’s place