JEE Main 2024 (31 Jan Shift 2)

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Animated Solution for Mathematics - Permutations and Combinations: If for some $m, n$ ; $^{6} C_{m} + 2 (^{6} C_{m + 1}) +^{6} C_{m + 2} >^{8} C_{3}$ and $^{n - 1} P_{3} :^{n} P_{4} = 1 : 8$ , then $^{n} P_{m + 1} +^{n + 1} C_{m}$ is equal to

Q: How do I know $m=2$ is the only solution for $^8C_{m+2} > 56$?

Great question! Remember that binomial coefficients $^nC_r$ increase until the middle term and then decrease. Since $^8C_3 = 56$ and $^8C_4 = 70$, the only value greater than 56 is 70. Thus, $m+2$ must be 4, which forces $m=2$.

Select Answer:

Progressive Hints

Hint 1

Look at the left-hand side of the inequality. It is a sum of three binomial coefficients. Does it remind you of the fundamental property of Pascal's Triangle? Specifically, how can you split the middle term

2 (^{6} C_{m + 1})

to create two pairs that each satisfy the identity

^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}

Hint 2

For the permutation ratio, resist the urge to expand everything into massive factorials immediately. Write out the definition of

^{n} P_{r} = \frac{n !}{( n - r )!}

for both the numerator and denominator. You will find that the

(n - 4)!

terms are identical and waiting to be cancelled.

Hint 3

To start the inequality, rewrite the expression as

(^{6} C_{m} +^{6} C_{m + 1}) + (^{6} C_{m + 1} +^{6} C_{m + 2})

. Once you apply Pascal's Identity to both pairs, you will be left with

^{7} C_{m + 1} +^{7} C_{m + 2}

. Apply the identity one more time to reach the final form

^{8} C_{m + 2} >^{8} C_{3}

Visualized Solution

Introduction to the Problem

Given Inequality: $^{6} C_{m} + 2 (^{6} C_{m + 1}) +^{6} C_{m + 2} >^{8} C_{3}$
Given Ratio: $^{n - 1} P_{3} :^{n} P_{4} = 1 : 8$
Objective: Find $^{n} P_{m + 1} +^{n + 1} C_{m}$

Analyzing the Combination Sum

LHS: $^{6} C_{m} + 2 (^{6} C_{m + 1}) +^{6} C_{m + 2}$
Split the middle term: $(^{6} C_{m} +^{6} C_{m + 1}) + (^{6} C_{m + 1} +^{6} C_{m + 2})$

Applying Pascal's Identity

Recall Pascal's Identity: $^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}$
First pair: $^{6} C_{m} +^{6} C_{m + 1} =^{7} C_{m + 1}$
Second pair: $^{6} C_{m + 1} +^{6} C_{m + 2} =^{7} C_{m + 2}$

Simplifying to a Single Term

Current expression: $^{7} C_{m + 1} +^{7} C_{m + 2}$
Apply Pascal's Identity again: $^{7} C_{m + 1} +^{7} C_{m + 2} =^{8} C_{m + 2}$

Evaluating the Inequality

Inequality becomes: $^{8} C_{m + 2} >^{8} C_{3}$
Calculate $^{8} C_{3}$ : $\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
Condition: $^{8} C_{m + 2} > 56$

Finding the Value of $m$

Values of $^{8} C_{r}$ peak at $r = 4$
For $^{8} C_{m + 2} > 56$ , $^{8} C_{m + 2}$ must be $70$
Therefore, $m + 2 = 4 ⟹ m = 2$

Analyzing the Permutation Ratio

Given: $\frac{^{n - 1} P _{3}}{^{n} P _{4}} = \frac{1}{8}$
Formula: $^{n} P_{r} = \frac{n !}{( n - r )!}$

Substituting the Formulas

Numerator: $^{n - 1} P_{3} = \frac{( n - 1 )!}{( n - 4 )!}$
Denominator: $^{n} P_{4} = \frac{n !}{( n - 4 )!}$
Equation: $\frac{\frac{( n - 1 )!}{( n - 4 )!}}{\frac{n !}{( n - 4 )!}} = \frac{1}{8}$

Simplifying the Factorials

Cancel $(n - 4)!$ : $\frac{( n - 1 )!}{n !} = \frac{1}{8}$
Expand $n!$ : $\frac{( n - 1 )!}{n ( n - 1 )!} = \frac{1}{8}$
Result: $\frac{1}{n} = \frac{1}{8} ⟹ n = 8$

Setting up the Final Expression

Expression: $^{n} P_{m + 1} +^{n + 1} C_{m}$
Substitute $n = 8, m = 2$ : $^{8} P_{2 + 1} +^{8 + 1} C_{2}$
Simplified Expression: $^{8} P_{3} +^{9} C_{2}$

Calculating the Final Terms

Calculate $^{8} P_{3}$ : $8 \times 7 \times 6 = 336$
Calculate $^{9} C_{2}$ : $\frac{9 \times 8}{2 \times 1} = 36$
Final Sum: $336 + 36 = 372$

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The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Binomial Inequality

We begin with the given inequality:

^{6} C_{m} + 2 (^{6} C_{m + 1}) +^{6} C_{m + 2} >^{8} C_{3}

To simplify the left-hand side, we split the middle term

2 (^{6} C_{m + 1})

into

^{6} C_{m + 1} +^{6} C_{m + 1}

. This allows us to group the terms as follows:

(^{6} C_{m} +^{6} C_{m + 1}) + (^{6} C_{m + 1} +^{6} C_{m + 2}) >^{8} C_{3}

Using Pascal's Identity, which states that

^{n} C_{r} +^{n} C_{r + 1} =^{n + 1} C_{r + 1}

, we simplify the grouped terms:

^{7} C_{m + 1} +^{7} C_{m + 2} >^{8} C_{3}

Applying Pascal's Identity once more, the expression collapses into a single term:

^{8} C_{m + 2} >^{8} C_{3}

Solving for $m$

First, we calculate the value of the constant on the right:

^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

We now require

^{8} C_{m + 2} > 56

. Examining the binomial coefficients for

n = 8

, we know the values increase until the middle term

^{8} C_{4} = 70

Since

70

is the only value in the row

n = 8

that is strictly greater than

56

, we must have:

m + 2 = 4 \Rightarrow m = 2

Solving for $n$

Next, we address the ratio condition:

^{n - 1} P_{3} :^{n} P_{4} = 1 : 8

Expanding this using the definition of permutations, we have:

\frac{( n - 1 )!}{( n - 1 - 3 )!} \div \frac{n !}{( n - 4 )!} = \frac{1}{8}

\frac{( n - 1 )!}{( n - 4 )!} \times \frac{( n - 4 )!}{n !} = \frac{1}{8}

The

(n - 4)!

terms cancel out, leaving:

\frac{( n - 1 )!}{n !} = \frac{1}{8}

Since

n! = n \times (n - 1)!

, the expression simplifies to:

\frac{1}{n} = \frac{1}{8} \Rightarrow n = 8

Final Calculation

We are tasked with evaluating the expression

^{n} P_{m + 1} +^{n + 1} C_{m}

using our derived values

m = 2

and

n = 8

^{8} P_{2 + 1} +^{9} C_{2} =^{8} P_{3} +^{9} C_{2}

Calculating the individual components:

^{8} P_{3} = 8 \times 7 \times 6 = 336

^{9} C_{2} = \frac{9 \times 8}{2} = 36

Adding these results together, we reach the final destination:

336 + 36 = 372

Common Traps & FAQs

Why can't I just expand

^{6} C_{m}

into factorials? It seems more direct.

While expansion works, it is a trap! It leads to complex algebraic expressions that are prone to calculation errors. The beauty of combinatorics lies in recognizing patterns like Pascal's Identity, which collapses the expression elegantly. Always look for the pattern before you reach for the brute-force expansion.

How do I know

m = 2

is the only solution for

^{8} C_{m + 2} > 56

Great question! Remember that binomial coefficients

^{n} C_{r}

increase until the middle term and then decrease. Since

^{8} C_{3} = 56

and

^{8} C_{4} = 70

, the only value greater than 56 is 70. Thus,

m + 2

must be 4, which forces

m = 2

Why does the

(n - 4)!

term cancel out so perfectly?

It is not a coincidence! The definition of permutations is designed to create these cancellations. When you divide

^{n} P_{4}

^{n - 1} P_{3}

, the

(n - 4)!

terms are common factors in the denominators of the factorial definitions, allowing them to vanish. This is the elegance of the factorial system at work.

Animated Solution for Mathematics - Permutations and Combinations: If for some $m, n$ ; $^{6} C_{m} + 2 (^{6} C_{m + 1}) +^{6} C_{m + 2} >^{8} C_{3}$ and $^{n - 1} P_{3} :^{n} P_{4} = 1 : 8$ , then $^{n} P_{m + 1} +^{n + 1} C_{m}$ is equal to

Select Answer:

Visualized Solution

Introduction to the Problem

Analyzing the Combination Sum

Applying Pascal's Identity

Simplifying to a Single Term

Evaluating the Inequality

Finding the Value of $m$

Analyzing the Permutation Ratio

Substituting the Formulas

Simplifying the Factorials

Setting up the Final Expression

Calculating the Final Terms

The Sigma Insight: Combinations and Selection

Analyzing the Binomial Inequality

Solving for $m$

Solving for $n$

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Introduction to the Problem

Analyzing the Combination Sum

Applying Pascal's Identity

Simplifying to a Single Term

Evaluating the Inequality

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Analyzing the Permutation Ratio

Substituting the Formulas

Simplifying the Factorials

Setting up the Final Expression

Calculating the Final Terms

The Sigma Insight: Combinations and Selection

Analyzing the Binomial Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Final Calculation

Similar Questions

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Finding the Value of $m$

Solving for $m$

Solving for $n$

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If $^{n - 1} C_{r} = (k^{2} - 3)^{n} C_{r + 1}$ , then $k \in$

If $^{n} P_{r} =^{n} P_{r + 1}$ and $^{n} C_{r} =^{n} C_{r - 1}$ , then the value of $r$ is equal to:

Let $S = {1, 2, 3, 4, 5, 6, 9}$ . Then the number of elements in the set $T = {A \subseteq S : A \neq = ϕ and the sum of all the elements of A is not a multiple of 3}$ is .

If $^{n} C_{r}$ denotes the number of combination of $n$ things taken $r$ at a time, then the expression $^{n} C_{r + 1} +^{n} C_{r - 1} + 2 \times^{n} C_{r}$ equals