Is there a faster way to calculate $\binom{10}{6}$?

Absolutely! Always look for symmetry. The property $\binom{n}{r} = \binom{n}{n-r}$ is a lifesaver. Here, $\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4}$. Since we already calculated $\binom{10}{4}$ as 210, we don't need to do the heavy lifting again. Always use these properties to save precious time in the exam!

JEE Main 2023 (08 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sets and Relations: Let the number of elements in sets $A$ and $B$ be five and two respectively. Then the number of subsets of $A \times B$ each having at least 3 and at most 6 elements is

Q: Wait, why is the total number of elements in $A \times B$ equal to 10?

Great question! The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. For every element in $A$, there are 2 corresponding elements in $B$. Since there are 5 elements in $A$, we have $5 \times 2 = 10$ unique pairs. Think of it as a grid with 5 rows and 2 columns; the total number of cells is the product of the dimensions.

Select Answer:

Visualized Solution

Given Sets $A$ and $B$

Given: $n (A) = 5$
Given: $n (B) = 2$

Cardinality of $A \times B$

$n (A \times B) = n (A) \times n (B)$
Substituting values: $5 \times 2 = 10$

Defining the Objective

Required: Number of subsets with $k$ elements
Condition: $3 \leq k \leq 6$
Total = $(3 10) + (4 10) + (5 10) + (6 10)$

Calculating $(3 10)$

Case 1: $k = 3$
$(3 10) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$
$(3 10) = 120$

Calculating $(4 10)$

Case 2: $k = 4$
$(4 10) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$
$(4 10) = 210$

Calculating $(5 10)$

Case 3: $k = 5$
$(5 10) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
$(5 10) = 252$

Calculating $(6 10)$

Case 4: $k = 6$
Using property: $(r n) = (n - r n)$
$(6 10) = (4 10) = 210$

Final Summation

Total = $120 + 210 + 252 + 210$
Total = $792$

Conclusion

Key Takeaway: Sum the individual combinations $(k n)$ for a given range.
Final Answer: $792$

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The Sigma Insight: Cartesian Product of Sets

Solution Diagram

Analyzing the Setup

We are given two sets,

A

and

B

, with cardinalities

n (A) = 5

and

n (B) = 2

The Cartesian product

A \times B

consists of all possible ordered pairs

(a, b)

where

a \in A

and

b \in B

. The total number of elements in this set is:

n (A \times B) = n (A) \times n (B) = 5 \times 2 = 10

The Master Equation

We need to find the number of subsets of

A \times B

that contain at least 3 and at most 6 elements. This is equivalent to calculating the sum of combinations:

k = 3 \sum 6 (k 10) = (3 10) + (4 10) + (5 10) + (6 10)

Step-by-Step Calculation

First, we calculate the number of ways to choose 3 elements:

(3 10) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

Next, we calculate the number of ways to choose 4 elements:

(4 10) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

For the case of 5 elements, we calculate:

(5 10) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

Finally, for

k = 6

, we utilize the symmetry property

(r n) = (n - r n)

(6 10) = (4 10) = 210

Final Calculation

Summing these individual results gives us the total number of valid subsets:

120 + 210 + 252 + 210 = 792

The total number of subsets with at least 3 and at most 6 elements is 792.

Common Traps & FAQs

Wait, why is the total number of elements in

A \times B

equal to 10?

Great question! The Cartesian product

A \times B

is defined as the set of all ordered pairs

(a, b)

where

a \in A

and

b \in B

. For every element in

A

, there are 2 corresponding elements in

B

. Since there are 5 elements in

A

, we have

5 \times 2 = 10

unique pairs. Think of it as a grid with 5 rows and 2 columns; the total number of cells is the product of the dimensions.

Why do we use combinations

(k n)

instead of permutations?

Remember that a subset is an unordered collection of elements. If you pick elements {x, y, z}, it is the same set as {z, x, y}. Since the order of elements within the subset does not matter, we use combinations, which count the number of ways to choose

k

items from

n

without regard to order.

Is there a faster way to calculate

(6 10)

Absolutely! Always look for symmetry. The property

(r n) = (n - r n)

is a lifesaver. Here,

(6 10) = (10 - 6 10) = (4 10)

. Since we already calculated

(4 10)

as 210, we don't need to do the heavy lifting again. Always use these properties to save precious time in the exam!

Animated Solution for Mathematics - Sets and Relations: Let the number of elements in sets $A$ and $B$ be five and two respectively. Then the number of subsets of $A \times B$ each having at least 3 and at most 6 elements is

Select Answer:

Visualized Solution

Given Sets $A$ and $B$

Cardinality of $A \times B$

Defining the Objective

Calculating $(3 10)$

Calculating $(4 10)$

Calculating $(5 10)$

Calculating $(6 10)$

Final Summation

Conclusion

The Sigma Insight: Cartesian Product of Sets

Analyzing the Setup

The Master Equation

Step-by-Step Calculation

Final Calculation

Similar Questions

Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of $A \times B$ having 3 or more elements is

Let $Z$ be the set of integers. If $A = {x \in Z : 2^{(x + 2) (x^{2} - 5 x + 6)} = 1}$ and $B = {x \in Z : - 3 < 2 x - 1 < 9}$ , then the number of subsets of the set $A \times B$ , is :

Let $A = {1, 3, 4, 6, 9}$ and $B = {2, 4, 5, 8, 10}$ . Let $R$ be a relation defined on $A \times B$ such that $R = {((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2} and b_{1} \leq a_{2}}$ . Then the number of elements in the set $R$ is

Let $A = {2, 3, 4}$ and $B = {8, 9, 12}$ . Then the number of elements in the relation $R = {((a_{1}, b_{1}), (a_{2}, b_{2})) \in (A \times B, A \times B) : a_{1} divides b_{2} and a_{2} divides b_{1}}$ is

Comprehension Passage

If $n (X) =^{m} C_{6}$ , then the value of $m$ is ________.

If the value of $n (Y) + n (Z)$ is $k^{2}$ , then $∣ k ∣$ is ________.

Which of the following options is the only INCORRECT combination ?

Select Answer:

Visualized Solution

Given Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Cardinality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A×B

Defining the Objective

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (310​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (410​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (510​)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (610​)

Final Summation

Conclusion

The Sigma Insight: Cartesian Product of Sets

Analyzing the Setup

The Master Equation

Step-by-Step Calculation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of A×B having 3 or more elements is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let Z be the set of integers. If A={x∈Z:2(x+2)(x2−5x+6)=1} and B={x∈Z:−3<2x−1<9}, then the number of subsets of the set A×B, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={2,3,4} and B={8,9,12}. Then the number of elements in the relation R={((a1​,b1​),(a2​,b2​))∈(A×B,A×B):a1​ divides b2​ and a2​ divides b1​} is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If n(X)=mC6​, then the value of m is ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the value of n(Y)+n(Z) is k2, then ∣k∣ is ________.

Which of the following options is the only INCORRECT combination ?

Given Sets $A$ and $B$

Cardinality of $A \times B$

Calculating $(3 10)$

Calculating $(4 10)$

Calculating $(5 10)$

Calculating $(6 10)$

Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of $A \times B$ having 3 or more elements is

Let $Z$ be the set of integers. If $A = {x \in Z : 2^{(x + 2) (x^{2} - 5 x + 6)} = 1}$ and $B = {x \in Z : - 3 < 2 x - 1 < 9}$ , then the number of subsets of the set $A \times B$ , is :

Let $A = {2, 3, 4}$ and $B = {8, 9, 12}$ . Then the number of elements in the relation $R = {((a_{1}, b_{1}), (a_{2}, b_{2})) \in (A \times B, A \times B) : a_{1} divides b_{2} and a_{2} divides b_{1}}$ is

If $n (X) =^{m} C_{6}$ , then the value of $m$ is ________.

If the value of $n (Y) + n (Z)$ is $k^{2}$ , then $∣ k ∣$ is ________.