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Concept:

• ⁿC_r = ⁿC_n-r.
• If nCx = nCy, then x = y or x + y = n.
• nCr + nCr-1 = n+1Cr.

 

Calculations:

\(\rm \begin{pmatrix} 15\ 8\end{pmatrix}+\begin{pmatrix} 15\ 7\end{pmatrix}=\begin{pmatrix} \rm n\ \rm r\end{pmatrix}\)

⇒ 15C8 + 15C7 = nCr

Using nCr + nCr-1 = n+1Cr:

⇒ ​15C8 + 15C8-1 = ¹⁵⁺¹C₈ = nCr

⇒ ​16C8 = nCr

⇒ n = 16 and r = 8.

Concept:

Let A and B denote two sets of elements.

• n(A) and n(B) are the number of elements present in set A and B respectively.
• n(A ⋃ B) is the total number of elements present in either set A or B.
• n(A ⋂ B) is the number of elements present in both the sets A and B.
• n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B)

 

Calculations:

Let A be the set of students who read "Hitavad" and B the set of students who read "Hindustan".

From the given information:

n(A ⋃ B) = 50 - 10 = 40.

n(A) = 30.

n(B) = 35.

By using n(A ⋃ B) = n(A) + n(B) - n(A ⋂ B), we get:

40 = 30 + 35 - n(A ⋂ B)

⇒ n(A ⋂ B) = 25.

∴ The number of students who read both "Hitavad" and "Hindustan" newpapers is n(A ⋂ B) = 25.

Concept:

We know that a set of which all the elements are contained in another set is known as a subset

Calculations:

Given, X = {4ⁿ - 3n - 1: n ϵ N}, where N is the set of natural numbers so N =1, 2, 3, ....

After putting all these values of N, we will get the elements of X.

⇒ X = {0, 9, 54,....}

Similarly, Y = {9(n - 1) : n ϵ N}, where N is the set of natural numbers,

⇒ Y = {0, 9, 18, 27,....}

Here, all elements of X is contained in set Y. 

Hence, X is a subset of Y.

\(\rm X \subset Y\)

Hence, if x = {4n - 3n - 1: n ϵ N} and Y{9(n - 1) : n ϵ N}, where N is the set of natural numbers, then \(\rm X \subset Y\).

Concept:

• The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Calculations:

Given, A = {x, y, z}. 

The power set (or powerset) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set.

Powerset of A = {ϕ,{x}, {y}, {z}, {x, y}, {y, z}, {x, z},{x, y, z}}.

Hence, the number of subsets in the powerset of A is 8.

Concept:

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possibilities for happening of events A and B are:

• Either event A OR event B alone = m + n.
• Both event A AND event B together = m × n.
• A number is divisible by 2 if its units digit is either 0, 2, 4, 6 or 8.

 

Permutations/Arrangements:

• The number of ways in which r objects can be arranged in n places is ⁿP_r = \(\rm \dfrac{n!}{(n - r)!}\).
• The number of ways in which n objects can be arranged in n places is nPn = n!.
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

 

Calculations:

The number of letters in the word DELHI = 5.

Since, D has to be the first letter, the number of ways in which the first letter can be written is 1.

The remaining 4 letters can be written in any order/arrangement of the remaining 4 letters (ELHI).

∴ The number of ways of writing the 4 letters = 4! = 4 × 3 × 2 × 1 = 24.

∴ The total number of ways of writing the 5 letters = 1 × 24 = 24.

Concept:

\(\rm ^nC_r = \dfrac{n!}{r!(n-r)!}\)

Calculations:

Given, Naresh has 10 friends, and he wants to invite 6 of them to a party.

Remove the 3 particular friends and invite 6 friends from the remaining = 10 - 3 = 7 friends.

This can be done in ⁷C₆ ways

 

Hence, required number of ways = 7C6 

⇒required number of ways = \(\rm \dfrac{7!}{(7-6)!6!}\)

⇒required number of ways = 7 

Hence, Naresh has 10 friends, and he wants to invite 6 of them to a party. total number of ways that 3 particular friends never attend the party = 7

Concept:

• The total number of ways in which n objects can be distributed in r people if any person can get 'any number of objects' = ^n+r-1C_r-1.
• The total number of ways in which n objects can be distributed in r people if every person must get 'at least one object' = ^n-1C_r-1.
• ⁿC_r = \(\rm \dfrac{n!}{r!(n-r)!}\).
• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

 

Calculations:

Here, the number of objects is n = 10 and the number of people is r = 4.

Since, every person must get at least one object, the total number of possible distributions is:

n-1Cr-1 = 10-1C4-1 = 9C3 

= \(\rm \dfrac{9!}{3!(9-3)!}=\dfrac{7\times8\times9}{3\times2\times1}\) 

= 84.

Concept:

• If the 'Speed' is constant, then the 'Distance' and the 'Time' are directly proportional to each other. It means that, if one of the quantities becomes n times, the other will also become n times.
• Speed = \(\rm \dfrac{Distance}{Time}\).
• Average Speed = \(\rm \dfrac{Total\ Distance}{Total\ Time}\).

 

Calculations:

Let's say that the distance between each pair of cities is 600 km.

From A to B:

Distance = 600 km.

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{600}{30}\) = 20 hrs.

From B to C:

Distance = 600 km.

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{600}{40}\) = 15 hrs.

From C to A:

Distance = 600 km.

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{600}{50}\) = 12 hrs.

For overall journey:

Total distance = 600 + 600 + 600 = 1800 km.

Total time = 20 + 15 + 12 = 47 hrs.

Average Speed = \(\rm \dfrac{Total\ Distance}{Total\ Time}=\dfrac{1800}{47}\) = 38.29 km/hr.

Alternate Method 

Formula used:

Speed = \(\rm \dfrac{Distance}{Time}\).

Average Speed = \(\rm \dfrac{Total\ Distance}{Total\ Time}\).

Calculation:

Let's say that the distance between each pair of cities be x km

According to the question,

Distance = x km.

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{x}{30}\) 

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{x}{40}\) 

Time = \(\rm \dfrac{Distance}{Speed}=\dfrac{x}{50}\)

Total distance = x/30 + x/40 + x/50 = 47x/600

Average Speed = Total distance/Total time = 3x × 600/47x = 38.29 km/hr.

Concept:

• P(not E) = 1 - P(E).
• Probability of a Compound Event [(A and B) or (B and C)] is calculated as:

P[(A and B) or (B and C)] = [P(A) × P(B)] + [P(C) × P(D)]

('and' means '×' and 'or' means '+')

 

Calculations:

It is given that P(A) = \(\dfrac{1}{2}\) and P(not B) = \(\dfrac{1}{4}\). Let's say that the probability of C solving the problem is P(C) = x.

∴ Probability of A not solving the problem = P(not A) = 1 - P(A) = \(1-\dfrac{1}{2}\) = \(\dfrac{1}{2}\).

And, probability of C not solving the problem = P(not C) = 1 - P(C) = 1 - x.

Now, the probability that the problem is not solved at all:

= P(not A) AND P(not B) AND P(not C)

= P(not A) × P(not B) × P(not C)

= \(\rm \dfrac{1}{2}\times\dfrac{1}{4}\times(1-x)\)

= \(\rm \dfrac{1-x}{8}\).

And, the probability that the problem is solved = 1 - Probability that the problem is not solved at all = \(\rm \dfrac{63}{64}\).

⇒ \(\rm 1-\dfrac{1-x}{8}=\rm \dfrac{63}{64}\)

⇒ \(\rm \dfrac{1-x}{8}=1-\rm \dfrac{63}{64}=\dfrac{1}{64}\)

⇒ \(\rm 1-x=\dfrac{1}{8}\)

⇒ \(\rm x=1-\dfrac{1}{8}=\dfrac78\).

∴ The probability of C solving the problem is P(C) = x = \(\dfrac78\).

Concept:

The probability that they will not win a prize in a single trial = 1 - The probability that they will win a prize in a single trial

Calculations:

Given, A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both win a prize.

Total number of possibilities = \(\rm 25 \times 25\)

Favourable number of their wining = 25

⇒ Probability of winning the prize = \(\dfrac {1}{25}\)

The probability that they will not win a prize in a single trial = 1 - The probability that they will win a prize in a single trial

The probability that they will not win a prize in a single trial = 1 - \(\dfrac {1}{25}\)

⇒The probability that they will not win a prize in a single trial = \(\dfrac {24}{25}\)