Permutations and Combinations - Online Test

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

To form a triangle 3 non collinear points are needed.

Number of ways of selecting 3  points out of 6 can be  done in 6C3 =6!3!.3!=1.2.3.4.5.6.1.2.3.1.2.3= 20 ways.   [∵nCr=n!r!(n−r)!]

If there are 10 things and we have  to make them in to two groups containing 6 things and 4 things respectively , you have to select 6 to form first group , then automatically another group would have formed of 4 remaining things.

Now 6 things can be selected from 10 things in 10C6 different ways

Also we have 10C6=10C4  [∵nCr=nCn−r]

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

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

Since they are boarding from ground floor and we are considering the number of ways they leave the lift ,we can consider there are 7 floor sas we exclude the ground floor)

As each of the 5 persons can leave the lift in 7 ways, required number of ways= 75 

We have  nCr - 1  + nCr = n+1Cr

Hence 12C4 + 12C5 = 13C5 ................(i)

But given 12C4 + 12C5 =nC5 ................(ii)

Comparing (i) and (ii) we get  n = 13

Total number of beads =11

We have 6 beads are alike and next 5 beads are also alike , also since it is a necklace it can be observed from both the sides .

Therefore required number of ways=(11−1)!2!.5!.6!=10!2!.5!.6!