Test:- Permutations And Combinations - 9 - Mathematics MCQ

Solve by simplifying the ratio

Solve by permuting 2 and 4 and making the equation out of it.

∴ 

r = 5 & n - r = 8 ⇒ n = 13

As 0 < x < 1, we have p < q.
The number of rational numbers = 5 + 4 + 3 + 2 + 1 = 15
When p, q have a common factor, we get some rational numbers which are not different from those already counted. There are 4 such numbers:

∴ The required number of rational numbers
= 15-4 = 11

The first place from the left can' be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in ⁹P₈ ways.
∴ the number of 9 -digit numbers = 9 x ⁹P₈. 

 x | x | x | Crosses can be filled in ³P₃ - ²P₂, ways
(∴ 0 cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers = (³P₃-²P₂) x 3!

Consider those 3 oersons as a single unit first, 6! 3!

As the order of A, B, C is not to change they are to be treated identical in arrangement. So, the required number of ways = 7!/3!

The number of numbers with 4 in the middle = ⁴P₄-³P₃
(∴ the other four places are to be filled by 0, 1,2 and 3, and a number cannot begin with 0).
Similarly, the number of numbers with 5 in the middle

The numbers are made of 1, 2, 3, 4, 5 or 0, 1 ,2 , 4 , 5.
∴ the required number of 5 digit numbers = 5!+(⁵P₅-⁴P₄).

A - {2 , 3 , 5, 7 , 11, 13, 17, 19, 23 , 29}. A rational number is made by taking any two in any order.
∴ the required number of rational numbers = ¹⁰P₂+ 1(including 1).

'-' signs will be put between two '+’ signs or at the two ends.
There arc 7 places for four '-' signs. So, the required number of ways

(there being no arrangement as the '+' signs are identical as well '-' as signs arc identical).

Arrange vowels and consonants separately

The units place can be filled in 2 ways.
The tens place can be filled in 3 ways.
The remaining two places can be filled by any two of the remaining three digits. So, the required number of numbers = 2 * 3 x ³P₂.

Each element of the set A can be given the image in the set A in k ways.
∴ the required number of functions,
i.e., n(S) = k x k x ... x (k times)= k^k.

Any selection of four digits from the ten digits 0, 1, 2, 3, ... 9 gives one number. So, the required number of numbers = ¹⁰C₄.

 The numbers will be made by 0, 2 , 4 , 6 or 0 , 4, 6, 8
∴ the required number of numbers 

The last place will be occupied by 0 or 2.