Test: Permutation, Combination & Probability - CAT MCQ

As the digits appear in ascending order in the numbers, number of ways of forming a n-digit number using the 9 digits = ⁹C_n
Number of possible two-digit numbers which can be formed
= ⁹C₂ + ⁹C₃ + ⁹C₄ + ⁹C₅ + ⁹C₆ + ⁹C₇ + ⁹C₈ + ⁹C₉
= (1 + 1)⁹ – (⁹C₀ + ⁹C₁) = 2⁹ – (⁹C₀ + ⁹C₁) = 512 – (1 + 9) = 502

If n is even, i.e., say n =  2 m then the number of ways is 2 × m! × m!, i.e., m odd numbers in alternate places and m even numbers in alternate places.
If n is odd, i.e., say n = 2m + 1, then the number of ways = m!(m + 1)!
Hence, either 2(m!)² = 72 or m! (m + 1)! = 72 If 2(m!)² = 72 ⇒ m! = 6 ⇒ m = 3
for m!(m + 1)! = 72, there is no solution.
Hence m = 3, and n = 2m = 6.

Letters of the word 'KAKA' can be arranged in

If we will take all vowels are together, then arrangements are
 3 arrangements.
So, (6 – 3) = 3 arrangements, where vowels are not together.
So, required probability = 

Total number of the cases = ⁶C₃ = 20.
The favourable cases for B surviving are:

Required  probability = 6/20 = 0.3.

First let the 6 other students be seated in 6 chairs.
The number of spaces between the 6 students = 7.
∴ Amar, Akbar and Anthony can be seated in the 7 places in ⁷C₃ ways.
Thus, the number of ways in the class photo can be taken such that no two of Amar, Akbar and Anthony are sitting together is
= ⁷C₃ × 3! × 6! = 151200.

p + q = 1, i.e. q = 1 – p (0 = p, q = 1)
Now when the sum of two variables is a constant then their multiplication is the maximum when they are equal.
So, pq will be maximum and   the
minimum when p = q = 1/2.
Thus, the minimum value of X  = 0.25 + 4 = 4.25.

If there are 60 chairs around a circular table, consider a scenario wherein there ae two chairs vacant between every two consecutive people. Thus, there will be exactly 20 people sitting and exactly 40 vacant seats between them and in such a scenario, next person coming to sit will have to sit next to someone.

Since the number of candies received by each child must get at least 2 candies.
Once each child has received 2 candies, the remaining 2 candies should be distributed in such a maniner that the number of candies with any child after the distribution remains a prime number.
The remaining 2 candies are given to exactly two children in such a way that both the children receive one candy each.
Hence, the number of ways of distribution = ⁸C₂ = 28.

An integer can end with any of ten digits = (0, 1, 2, ......9) but (0, 1, 5, 6) has same unit of ten product of two integers.
So, the probability of an integer ending with 0 or 1 or 5 or 6.

and the probability of 2nd integer = 1/10
∴  Required probability = 

The number of yellow balls initially in the box is the same as the number of green balls. None of the ten operations involved after this favour any particular colour between the two. So the probability of the final ball picked being yellow (p(Y)) must be the same as that of being green (p(G)).
Thus, p(Y) = p(G) and the ball picked should be either yellow or green, which means that p(Y) + p(G) = 1.
Hence, p(Y) = p(G) = 1/2