Test: Permutation & Combination- 2 - CAT MCQ

The total number of ways to select 4 children out of 10 total children is (104)=10∗9∗8∗74∗3∗2∗1=210. However, because at least one boy must be in the group, we have to subtract 1 to account for the case where all four children are girls (There is (44)=1 such case). The answer is 210−1=209.

Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400

Total number of different numbers of 4 digits formed by using these digits only once = 4!/2! = 12.

The numbers formed whose unit digit is 3 = 3

The number formed whose unit digit is 5 = 6

The number formed whose unit digit is 6 = 3

(Same rule is applicable for 10th, 100th and 1000th units)

Sum of the total number of unit digits = 3×3+5×6 +6×3 = 57

Sum of the 10th digit number = 57×10

Sum of the 100th digit number = 57×100

Sum of the 1000th digit number = 57×1000

Sum of all the numbers formed = 57(1+10+100 + 1000) = 63327.

So, India can win the series in 1+4+10+20 = 35 ways.
Similarly, Australia can win the series in 35 ways.
Hence, the total number of ways in which the series can be won is 35+35=70 ways.

The digits which can be recognised as unique digits when they are Inverted in a calculator are 0, 1, 2, 5, 6, and 9 Since the number cannot begin with zero all the numbers having 0 at units place should be discarded for otherwise when reed upside down the number yen begin with 3 we now :1St the different posibililities

Thus, the number of required numbers = 7 + 6² + 6² x 7 + .. + 6² x 7⁴ 

Divide the chess board into two parts by a diagonal line 80 To select 4 squares lying on
diagonal line we have to choose the squares lying on P₁, Q₁, P₂Q₂, P₃, Q₃, P₄Q₄ and BD.
Number of squares on one side of BD =⁴C₄ +⁵C₄ +⁶C₄ + ⁷C₄ = 56. Thus total squares on either side of BO = 56 x 2 = 112.
On the diagonal BD, we will have 8C4, squares=70. So total squares = 112 . 70 = 182.
Similarly on the other side of the diagonal line AC we will have 182 squares.
Hence total number of squares = 2 x 182 = 364

4 letters can go into 4 envelopes in 4! Ways, of these onIy one choice is the correct one.

So. total number of wrong choices = 4! - 1

There are 11 letters in the word 'PATALIPUTRA‘ and there are two P's. two T's. three A's and four other different letters. Number of consonants = 6. number of vowels = 5.

Since relative order of the vowels and consonants remains unchanged. vowels can occupy only vowel’s place and consonants can occupy only consonants place.

Now 6 consonants can be arranged among themselves 6!/2!2! ways (since there are two P's and two T’s).

And 5 vowels can be arranged among themselves in 5!/3!ways (since A occurs thrice)

Therefore, Required number = 6!/2!2! x 5!/3! = 3600.