Permutations and Combinations Test - 3

We have a regular tetrahedron has 4 faces and we have to colour it with 4 different colours in 4! = 24 ways.

But in this we will be getting many overcountings .

We have there are 12 ways in which we can orient a regular tetrahedron  

Hence the number of distinct ways of colouring a regular tetrahedron  with 4 different colours is 24/12 = 2

1.Arrange all the alphabets in alphabetical order ( E, F , I  , W )

2. 4 alphabets can form 4! words = 24 words.

Since we need an even number , ones place can be occupied by only two numbers 2 and 4 in two ways.

Since repetition is not allowed tens place is occupied by remaining 4 numbers in 4 ways and the hundred's place by 3 ways and thousands place in 2 ways.

Hence total number of even numbers can be formed in 1x2x3x4x2 = 48.

Number of ways of forming 5 digit numbers= 5×4×3×2×1=120

 Number of ways of forming 4 digit numbers= 5×4×3×2=120

Number of ways of forming 3 digit numbers= 5×4×3=60

Number of ways of forming 2 digit numbers=5×4=20

Number of ways of forming 1 digit numbers= 5

Hence the total number of ways = 120+ 120 + 60+ 20+ 5 = 325

 Since given there are four alternatives in which one or more are correct,we have to consider the following four cases

The candidate choose 1 correct answer, 2 correct answers,3 correct answers or 4 correct answers.

1 correct answer can be chosen in ⁴C₁ ways = 4 ways

 2 correct answer can be chosen in ⁴C₂ ways= 6 ways

 3 correct answers can be chosen in  ⁴C₃ways  = 4 ways

4 correct answers can be chosen in⁴C₄ ways = 1 way

 Hence the totalnumber of ways = 4 + 6 + 4+ 1=15ways

In the word 'MATHAEMATICS'  there are   7 consonants which are M-2 ,,H-1,C-1,S-1, T-2 .Since they have to occur together we treat them as a single unit.

Now this single unit together with the remaining 4 vowels which are A-2,E-1,and I-1  will account for 5 letters.

Now in these 5 letters we have A is repeating twice  these can be aarranged in 5!/2!  different ways.

Corresponding to each of these arrangements the consonents can be arranged in 7!/2!.2! different ways.

Hence the number of ways it can be arranged is 7!5!/2!2!2! =75600

No: of combinations of n different things taken r at at a time is given by ⁿC_r

No: of combinations of n different things taken r at a time and excluding m particular things is ^{n - m}C_r = C(n-m,r)