JEE Advanced 2014

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is

Q: Why can't I just use the Stars and Bars formula? It's usually for sums like this!

That is a fantastic question! Stars and Bars is a powerful tool, but it assumes we are dealing with unordered, non-negative integer solutions where repetition is allowed. Here, the problem is fundamentally different: the order is fixed ($n_1 < n_2 < n_3 < n_4 < n_5$), and the integers must be distinct. The constraint is not just about the sum; it is about the structure of the set. Stars and Bars would overcount wildly because it doesn't respect the 'strictly increasing' rule.

Q: How do I know I haven't missed a case?

By fixing the largest variable, $n_5$, you create a partition of the problem into mutually exclusive cases. As long as you iterate through the possible range of $n_5$ (from $6$ to $10$), you are guaranteed to cover every possibility without overlap. It is like organizing a library: if you sort by the largest book first, you will never lose track of the smaller ones.

Q: Why is $n_5 = 5$ impossible?

Let's test the logic! If $n_5 = 5$, then the largest possible sum for the first four integers (using the smallest distinct positive integers $1, 2, 3, 4$) is $10$. Adding $n_5=5$ gives a total of $15$. Since $15 < 20$, it is mathematically impossible to reach the target sum of $20$ with $n_5=5$. The 'floor' of the sum is simply too high.

Enter Numerical Value:

Progressive Hints

Hint 1

Think of the constraint

n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

as a rigid staircase. You cannot simply pick numbers; they must climb. The most powerful way to tame this is to anchor the largest value,

n_{5}

, and see what 'budget' remains for the others. This transforms a chaotic search into a structured, manageable process.

Hint 2

Use the 'Minimum Sum' principle. Since the integers are distinct and positive, the smallest possible values for the first four are

1, 2, 3,

and

4

. Their sum is

10

. This immediately tells you that

n_{1} + n_{2} + n_{3} + n_{4}

must be at least

10

, which places a strict ceiling on

n_{5}

. If

n_{5}

were too small, you could never reach the total sum of

20

Hint 3

Start your systematic search by fixing

n_{5}

. Since

n_{1} + n_{2} + n_{3} + n_{4} = 20 - n_{5}

, and we know

n_{1} + n_{2} + n_{3} + n_{4} \geq 10

, you can deduce the range of

n_{5}

. Then, for each valid

n_{5}

, find the number of ways to partition the remaining sum into 4 distinct integers, all strictly less than

n_{5}

. This is the final step to unlocking the answer.

Visualized Solution

$U n d er s t an d in g t h e C o n s t r ain t s$

Given $5$ positive integers: $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}$
Sum condition: $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$

$T h e S t r i c tl y I n cr e a s in g C o n d i t i o n$

Condition: $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$
All numbers must be distinct and strictly increasing.

$E s t ab l i s hin g t h e L o w er B o u n d$

Minimum possible values for the first four integers:
$n_{1} \geq 1$ , $n_{2} \geq 2$ , $n_{3} \geq 3$ , $n_{4} \geq 4$
Minimum sum of first four: $1 + 2 + 3 + 4 = 10$

$M a x im u mV a l u e f or n_{5}$

Since $n_{1} + n_{2} + n_{3} + n_{4} \geq 10$
Maximum possible value for $n_{5} = 20 - 10 = 10$
So, $n_{5}$ can range from $6$ to $10$ .

$C a se 1 : n_{5} = 10$

If $n_{5} = 10$ , then $n_{1} + n_{2} + n_{3} + n_{4} = 10$
The only distinct positive integers summing to $10$ are $(1, 2, 3, 4)$ .
Solution 1: $(1, 2, 3, 4, 10)$

$C a se 2 : n_{5} = 9$

If $n_{5} = 9$ , then $n_{1} + n_{2} + n_{3} + n_{4} = 11$
Possible set: $(1, 2, 3, 5)$
Note: $(1, 2, 4, 4)$ is invalid (numbers must be distinct).
Solution 2: $(1, 2, 3, 5, 9)$

$C a se 3 : n_{5} = 8$

If $n_{5} = 8$ , then $n_{1} + n_{2} + n_{3} + n_{4} = 12$
Possible sets:
$1.$ $(1, 2, 3, 6)$
$2.$ $(1, 2, 4, 5)$
Solutions 3 & 4: $(1, 2, 3, 6, 8)$ and $(1, 2, 4, 5, 8)$

$C a se 4 : n_{5} = 7$

If $n_{5} = 7$ , then $n_{1} + n_{2} + n_{3} + n_{4} = 13$
Possible sets:
$1.$ $(1, 2, 4, 6)$
$2.$ $(1, 3, 4, 5)$
Invalid: $(1, 2, 3, 7)$ because $n_{4}$ must be less than $n_{5}$ .
Solutions 5 & 6: $(1, 2, 4, 6, 7)$ and $(1, 3, 4, 5, 7)$

$C a se 5 : n_{5} = 6$

If $n_{5} = 6$ , then $n_{1} + n_{2} + n_{3} + n_{4} = 14$
The only valid set with $n_{4} < 6$ is $(2, 3, 4, 5)$ .
Solution 7: $(2, 3, 4, 5, 6)$
Note: If $n_{5} = 5$ , max sum of first four is $1 + 2 + 3 + 4 = 10$ , total sum $15 < 20$ .

$F ina l T a l l y an d T ak e a w a y s$

Summary of solutions:
Case $n_{5} = 10$ : $1$ solution
Case $n_{5} = 9$ : $1$ solution
Case $n_{5} = 8$ : $2$ solutions
Case $n_{5} = 7$ : $2$ solutions
Case $n_{5} = 6$ : $1$ solution
Total number of arrangements = $7$

00:00 / 00:00

The Sigma Insight: Combinations and Selection

Solution Diagram

Analyzing the Setup

We are tasked with finding the number of distinct arrangements of five positive integers

(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})

such that they satisfy the strict inequality:

n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

Additionally, these integers must satisfy the sum constraint:

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20

This is a problem of partitions with distinct parts. Because the integers are strictly increasing, the largest value,

n_{5}

, acts as an anchor that limits the possible values for the preceding integers.

The Anchor Strategy

To determine the range of

n_{5}

, we consider the smallest possible values for the first four integers, which are

1, 2, 3,

and

4

. Their sum is

1 + 2 + 3 + 4 = 10

Since

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20

, and

n_{1} + n_{2} + n_{3} + n_{4} \geq 10

, it follows that

n_{5}

must satisfy:

n_{5} \leq 20 - 10 = 10

Conversely, if

n_{5} = 5

, the maximum possible sum is

1 + 2 + 3 + 4 + 5 = 15

, which is less than

20

. Therefore,

n_{5}

must be an integer in the range

6 \leq n_{5} \leq 10

The Case-by-Case Investigation

We now evaluate each possible value for

n_{5}

to find valid sets

(n_{1}, n_{2}, n_{3}, n_{4})

Case 1: $n_{5} = 10$ We require

n_{1} + n_{2} + n_{3} + n_{4} = 10

. The only set of four distinct positive integers summing to

10

(1, 2, 3, 4)

. Solution:

(1, 2, 3, 4, 10)

Case 2: $n_{5} = 9$ We require

n_{1} + n_{2} + n_{3} + n_{4} = 11

. The valid set is

(1, 2, 3, 5)

. Solution:

(1, 2, 3, 5, 9)

Case 3: $n_{5} = 8$ We require

n_{1} + n_{2} + n_{3} + n_{4} = 12

. The valid sets are

(1, 2, 3, 6)

and

(1, 2, 4, 5)

. Solutions:

(1, 2, 3, 6, 8)

and

(1, 2, 4, 5, 8)

Case 4: $n_{5} = 7$ We require

n_{1} + n_{2} + n_{3} + n_{4} = 13

. The valid sets are

(1, 2, 4, 6)

and

(1, 3, 4, 5)

. Solutions:

(1, 2, 4, 6, 7)

and

(1, 3, 4, 5, 7)

Case 5: $n_{5} = 6$ We require

n_{1} + n_{2} + n_{3} + n_{4} = 14

. The only valid set is

(2, 3, 4, 5)

. Solution:

(2, 3, 4, 5, 6)

Final Calculation

Summing the number of solutions found in each case: -

n_{5} = 10

1

solution -

n_{5} = 9

1

solution -

n_{5} = 8

2

solutions -

n_{5} = 7

2

solutions -

n_{5} = 6

1

solution

The total number of distinct arrangements is

1 + 1 + 2 + 2 + 1 = 7

Final Answer: 7

Common Traps & FAQs

Why can't I just use the Stars and Bars formula? It's usually for sums like this!

That is a fantastic question! Stars and Bars is a powerful tool, but it assumes we are dealing with unordered, non-negative integer solutions where repetition is allowed. Here, the problem is fundamentally different: the order is fixed (

n_{1} < n_{2} < n_{3} < n_{4} < n_{5}

), and the integers must be distinct. The constraint is not just about the sum; it is about the structure of the set. Stars and Bars would overcount wildly because it doesn't respect the 'strictly increasing' rule.

How do I know I haven't missed a case?

By fixing the largest variable,

n_{5}

, you create a partition of the problem into mutually exclusive cases. As long as you iterate through the possible range of

n_{5}

(from

6

10

), you are guaranteed to cover every possibility without overlap. It is like organizing a library: if you sort by the largest book first, you will never lose track of the smaller ones.

Why is

n_{5} = 5

impossible?

Let's test the logic! If

n_{5} = 5

, then the largest possible sum for the first four integers (using the smallest distinct positive integers

1, 2, 3, 4

) is

10

. Adding

n_{5} = 5

gives a total of

15

. Since

15 < 20

, it is mathematically impossible to reach the target sum of

20

with

n_{5} = 5

. The 'floor' of the sum is simply too high.

Animated Solution for Mathematics - Permutations and Combinations: Let $n_{1} < n_{2} < n_{3} < n_{4} < n_{5}$ be positive integers such that $n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 20$ . Then the number of such distinct arrangements $(n_{1}, n_{2}, n_{3}, n_{4}, n_{5})$ is

Enter Numerical Value:

Visualized Solution

$U n d er s t an d in g t h e C o n s t r ain t s$

$T h e S t r i c tl y I n cr e a s in g C o n d i t i o n$

$E s t ab l i s hin g t h e L o w er B o u n d$

$M a x im u mV a l u e f or n_{5}$

$C a se 1 : n_{5} = 10$

$C a se 2 : n_{5} = 9$

$C a se 3 : n_{5} = 8$

$C a se 4 : n_{5} = 7$

$C a se 5 : n_{5} = 6$

$F ina l T a l l y an d T ak e a w a y s$

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Anchor Strategy

The Case-by-Case Investigation

Final Calculation

Similar Questions

Let S denote the set of 4-digit numbers $ab c d$ such that $a > b > c > d$ and P denote the set of 5-digit numbers having product of its digits equal to 20. Then $n (S) + n (P)$ is equal to ......... .

Let $S = {1, 2, 3, ....., 9}$ . For $k = 1, 2, ....., 5$ , let $N_{k}$ be the number of subsets of S, each containing five elements out of which exactly $k$ are odd. Then $N_{1} + N_{2} + N_{3} + N_{4} + N_{5} =$

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

Let $n$ and $k$ be positive such that $n \geq \frac{k ( k + 1 )}{2}$ . The number of solutions $(x_{1}, x_{2}, \dots, x_{k})$ , $x_{1} \geq 1, x_{2} \geq 2, \dots, x_{k} \geq k$ , all integers, satisfying $x_{1} + x_{2} + \dots + x_{k} = n$ , is .........

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Five balls of different colours are to be placed in there boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty ?

Enter Numerical Value:

Visualized Solution

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

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case1:n5​=10

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case2:n5​=9

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case3:n5​=8

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case4:n5​=7

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case5:n5​=6

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

The Sigma Insight: Combinations and Selection

Analyzing the Setup

The Anchor Strategy

The Case-by-Case Investigation

Final Calculation

Similar Questions

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is _____

Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

Five balls of different colours are to be placed in there boxes of different size. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty ?

$U n d er s t an d in g t h e C o n s t r ain t s$

$T h e S t r i c tl y I n cr e a s in g C o n d i t i o n$

$E s t ab l i s hin g t h e L o w er B o u n d$

$M a x im u mV a l u e f or n_{5}$

$C a se 1 : n_{5} = 10$

$C a se 2 : n_{5} = 9$

$C a se 3 : n_{5} = 8$

$C a se 4 : n_{5} = 7$

$C a se 5 : n_{5} = 6$

$F ina l T a l l y an d T ak e a w a y s$