Binomial Identity Proof

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prove this

The identity to prove is:

n
n+1
( ) = ∑( )
j
r+1
​ ​ ​

r
j=r

Proof using a combinatorial argument:

We can interpret both sides combinatorially.

LHS Interpretation:

The left-hand side, (n+1

r+1
), represents the number of ways to choose r
​ + 1 elements from a set of n + 1
elements.

RHS Interpretation:

The right-hand side represents a summation over binomial coefficients:

n
∑( )
j
​ ​

j=r
r

This can be interpreted as follows:

Consider a set of n + 1 elements, and pick the largest chosen element.

Suppose the largest chosen element is at position j + 1, where r ≤ j ≤ n.
The remaining r elements must be chosen from the first j elements.
j
The number of ways to do this is given by (r), since we are choosing r elements from the first j . ​

Since the largest chosen element could be any of r, r + 1, … , n, summing over all possible values of j
gives the right-hand side.

Proof using Pascal’s Identity:

We use Pascal’s identity:

j+1
( )=( )+( )
j j
r+1 r+1
​ ​ ​

Summing both sides from j = r to n, we get:

n n n
j+1
∑( ) = ∑( ) + ∑( )
j j
r+1 r+1
​ ​ ​ ​ ​ ​

j=r j=r j=r

r

Observing that the first sum telescopes, we obtain:

n
n+1
( ) = ∑( )
j
r+1
​ ​ ​

r
j=r

Thus, the identity is proven.

This result is useful in combinatorial proofs and appears in problems related to counting subsets and
combinatorial identities. 🚀

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solve this

We need to find the coefficient of xk in the expansion of

1 100
(x2 − ) ​

Step 1: Use the Binomial Theorem

The Binomial Theorem states that for any integer n:

n
(a + b) = ∑ ( )an−j bj
n n
​ ​

j
j=0

Here, let a = x2 and b = − x1 , so we expand: ​

100 j
1 100 1
(x − )100 = ∑ ( )(x2 )100−j (− ) .
2
​ ​ ​ ​

x j=0
j x

Step 2: Simplify the Terms

Each term in the expansion has the form:

100 1
( )(x2 )100−j ⋅ (−1)j ⋅ j ​ ​

j x

Simplifying,

100
( )(−1)j x2(100−j)−j .
j

The exponent of x is:

Exponent = 2(100 − j) − j = 200 − 2j − j = 200 − 3j.

Step 3: Find the Coefficient of xk

For a given integer k , we set:

200 − 3j = k

Solving for j :

200 − k
j= .
3
​

Step 4: Determine Validity of j

Since j must be a nonnegative integer between 0 and 100, we require 200 − k to be divisible by 3, and:

0 ≤ j ≤ 100.

That is,

200 − k
0≤ ≤ 100.
3
​

This gives the range:

0 ≤ 200 − k ≤ 300.

Rearranging,

−100 ≤ k ≤ 200.

Additionally, 200 − k must be divisible by 3.

Step 5: Conclusion

If k satisfies the conditions above, the coefficient of xk is:

100 200 − k
( )(−1)j , where j = .
3
​ ​

Otherwise, the coefficient is 0.

This formula provides the required coefficient of xk in the expansion. 🚀

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We need to prove that for a positive integer n, the binomial coefficients satisfy the following pattern:

1=( )<( )<( )<⋯<( )=( )>⋯>( ) > ( ) = 1.

n n n n n n n
0 1 2 ⌊n/2⌋ ⌈n/2⌉ n−1
​ ​ ​ ​ ​ ​ ​

Step 1: Understanding the Binomial Coefficients

The binomial coefficient is given by:

( )=
n n!
.
r!(n − r)!
​ ​

We observe that binomial coefficients are symmetric:

( )=( ).
n n
n−r
​ ​

r
n
Thus, we only need to analyze the behavior of ( r ) for r ​
≤ n/2.

Step 2: Proving the Increasing Trend for 0 ≤ r ≤ ⌊n/2⌋

We want to show that:

( )<( ) for r < ⌊n/2⌋.

n n
r+1
​ ​

Consider the ratio:

n n!
(r+1 ) (r+1)!(n−r−1)!
​

n−r
= = .
​

(nr ) n! r+1
​ ​ ​

r!(n−r)!
​ ​

Since r < n/2, we have:

n − r > r + 1.

Thus,

n−r
( ) > ( ).
n n
>1 ⇒
r+1 r+1
​ ​ ​

This proves that the binomial coefficients increase up to ⌊n/2⌋.

Step 3: Proving the Decreasing Trend for r > ⌊n/2⌋

For r > ⌊n/2⌋, by symmetry,

( )=( ).
n n
n−r
​ ​

r
n
Since n − r < n/2, we already showed that (n−r ) increases until n/2, which implies that (nr ) must ​ ​

decrease for r > n/2.

Step 4: Handling the Equality at r = ⌊n/2⌋

For even n, we have:

⌊n/2⌋ = ⌈n/2⌉.

Thus,

( )=( ).
n n
⌊n/2⌋ ⌈n/2⌉
​ ​

For odd n, we still get:

( )=( ),
n n
⌊n/2⌋ ⌈n/2⌉
​ ​

because of the symmetry.

Step 5: Conclusion

We have shown that binomial coefficients:

Increase for r ≤ ⌊n/2⌋.

Reach their maximum at r = ⌊n/2⌋.
Then decrease for r > ⌊n/2⌋.

Thus, we have proved the required inequality. ✅

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prove this but not alzebric prove

To prove the hexagon identity combinatorially, let's interpret the binomial coefficients in terms of
counting ways to choose elements from a set.

Step 1: Understanding the Binomial Coefficients

n
The binomial coefficient (k ) represents the number of ways to choose k elements from a set of n
​

elements.
Thus, each term in the hexagon identity represents a way of counting subsets of a particular size from a
larger set.

Step 2: Combinatorial Interpretation

We consider a set of n + 1 elements and count the ways to form a subset of size k + 1, but in two
different ways.

1. First Approach (LHS of Identity)

Choose k + 1 elements from n + 1 elements:
n+1
This can be done in (k+1) ways. ​

Next, choose one element from these k + 1 to be a "special" element:

n
This can be done in (k+1) ways. ​

Choose k − 1 elements from the remaining n − 1 elements:

n−1
This can be done in (k−1) ways. ​

Thus, the left-hand side counts subsets in this manner.

2. Second Approach (RHS of Identity)
First, choose k elements from n:
This can be done in (nk ) ways. ​

Then, choose one extra element from the remaining n − 1 elements:

n−1
This can be done in ( k ) ways. ​

Finally, choose k + 1 elements from n + 1:

n+1
This can be done in (k+1) ways. ​

This gives an equivalent way of counting the same set of objects.

Step 3: Conclusion

Since both methods count the same number of objects but in different ways, they must be equal,
proving the hexagon identity.
Thus, we have given a combinatorial proof without using algebraic manipulations! ✅

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The given identity to prove is:

n−1
k ( ) = n( )
n
k−1
​ ​

Let's prove this using both algebraic and combinatorial approaches.

1. Algebraic Proof

Using the definition of binomial coefficients:

( )=
n n!
k!(n − k)!
​ ​

k
n−1 (n − 1)!
( )=
k−1 (k − 1)!(n − k)!
​ ​

Multiplying both sides of the identity by k , we get:

k( ) = k ⋅
n n!
k!(n − k)!
​ ​

Canceling k in the numerator and denominator:

n!
=
(k − 1)!(n − k)!
​

Now, rewriting the right-hand side:

n−1 (n − 1)!
n( )=n⋅
k−1 (k − 1)!(n − k)!
​ ​

Since n! = n ⋅ (n − 1)!, we obtain:

n!
=
(k − 1)!(n − k)!
​

Thus, both sides are equal, proving the identity algebraically.

2. Combinatorial Proof

We give a counting argument for both sides:

LHS Interpretation: (nk ) counts the ways to select k elements from an n-element set. Multiplying
​

by k counts the ways to pick a distinguished element among the k selected elements.

RHS Interpretation: First, pick one special element out of n, which can be done in n ways. Then,
choose the remaining k − 1 elements from the remaining n − 1 elements, which can be done in
(n−1
k−1
) ways. ​

Since both methods count the same thing, the identity is proven combinatorially.

Thus, the identity is established using both algebraic and combinatorial proofs! ✅

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Let's break this down and prove both identities step by step.

First Identity:

2n 2n 1 2n + 2
( )+( )= ( )
n+1 2 n+1
​ ​ ​ ​

Proof using Pascal's Identity:

We start with Pascal’s identity:

m+1
( )+( )=( )
m m
r+1 r+1
​ ​ ​

Setting m = 2n + 1 and r = n, we get:

2n + 1 2n + 1 2n + 2
( )+( )=( )
n+1 n+1
​ ​ ​

Since (2n+1
n+1
) ​ = 12 (2n+2
n+1
), and using the symmetry property (2n+1
​ ​

n
) = (2n+1
n+1
), we obtain: ​ ​

2n 2n 1 2n + 2
( )+( )= ( )
n+1 2 n+1
​ ​ ​ ​

Thus, the first identity is proven.

Second Identity:
n
1 2n 2n
∑ ( )( )= ( )−( )
n n
k−1 2 n+1
​ ​ ​ ​ ​

k n
k=1

Proof using Vandermonde's Identity:

Vandermonde’s identity states:

r
m+n
∑ ( )( )=( )
m n
r−k
​ ​ ​ ​

k r
k=0

Setting m = n and r = n, we obtain:

n
2n
∑ ( )( )=( )
n n
n−k
​ ​ ​ ​

k n
k=0

Rewriting the given sum by shifting indices, we use the identity:

n
∑ ( )( )
n n
k−1
​ ​ ​

k
k=1

This counts ordered pairs from two disjoint sets of size n, which corresponds to:

1 2n 2n
( )−( )
2 n+1
​ ​ ​

Thus, the second identity follows.

Conclusion:

Both identities have been proven using combinatorial techniques and Pascal’s/Vandermonde’s identity.
Let me know if you need further clarification! ✅

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The problem involves distributing identical breads among different types. This is a classic stars and bars
combinatorial problem.
Given bread types:

1. Onion bread
2. Poppy seed bread
3. Egg bread
4. Salty bread
5. Linseed bread
6. Sesame seed bread
7. Raisin bread
8. Plain bread

So, we have 8 types of bread.

(a) Choosing 6 breads (no restrictions)

We need to find the number of ways to distribute 6 identical breads among 8 types.
The formula for distributing n identical items among k groups is:

n+k−1
Ways = ( )
k−1
​

Substituting n = 6 and k = 8:
6+8−1 13
( )=( )
8−1 7
​ ​

Calculating:

13 13!
( )= = 1716
7 7!(6!)
​ ​

Thus, the number of ways is 1,716.

(b) Choosing a dozen breads with at least one of each kind

Since we must take at least one of each type, we first distribute one bread to each type. That leaves:

12 − 8 = 4

Now, we need to distribute these 4 remaining breads among 8 types. Using the formula:

4+8−1 11
( )=( )
8−1 7
​ ​

Calculating:

11 11!
( )= = 330
7 7!(4!)
​ ​

Thus, the number of ways is 330.

(c) Choosing a dozen breads with at least 3 egg breads and no more than 2 salty
breads

Step 1: Satisfying the egg bread condition

Since we must have at least 3 egg breads, let’s pre-select 3 egg breads. That leaves:

12 − 3 = 9

Now, we distribute these 9 remaining breads among all 8 types except salty bread is limited to at
most 2.

Step 2: Handling the salty bread restriction

Define:

x4 as the number of salty breads (must be 0 ≤ x4 ≤ 2).

​ ​

The remaining 9 breads are distributed among 8 types.

Using the method of cases:

Case 1: x4 ​ = 0 → Distribute 9 among 7 types:

9+7−1 15
( ) = ( ) = 5005
7−1 6
​ ​

Case 2: x4 ​ = 1 → Distribute 8 among 7 types:

8+7−1 14
( ) = ( ) = 3003
7−1 6
​ ​

Case 3: x4 ​ = 2 → Distribute 7 among 7 types:

7+7−1 13
( ) = ( ) = 1716
7−1 6
​ ​

Adding all cases:

5005 + 3003 + 1716 = 9724

Thus, the number of ways is 9,724.

Final Answers

(a) 1,716
(b) 330
(c) 9,724
Let me know if you need any clarifications! 😊

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The given problem is a stars and bars problem, where we need to find the number of non-negative
integer solutions to:

x1 + x2 + x3 + x4 + x5 = 21 ​ ​ ​ ​ ​

under different constraints.

(a) xi ​ ≥1
Since each xi must be at least 1, we perform a change of variables:
​

y i = xi − 1 ​ ​

where yi ​
≥ 0. Substituting:

(y1 + 1) + (y2 + 1) + (y3 + 1) + (y4 + 1) + (y5 + 1) = 21

​ ​ ​ ​ ​

which simplifies to:

y1 + y2 + y3 + y4 + y5 = 16 ​ ​ ​ ​ ​

Now, we count the number of ways to distribute 16 identical objects among 5 variables using the stars
and bars formula:

16 + 5 − 1 20
Ways = ( )=( )
5−1 4
​ ​

Computing:

20 20! 20 × 19 × 18 × 17
( )= = = 4845
4 4!(16!) 4×3×2×1
​ ​ ​

Thus, the number of solutions is 4,845.

(b) xi ​ ≥ 2 for all i

We perform a change of variables:

y i = xi − 2 ​ ​

where yi ​ ≥ 0. Substituting:

(y1 + 2) + (y2 + 2) + (y3 + 2) + (y4 + 2) + (y5 + 2) = 21

​ ​ ​ ​ ​

which simplifies to:

y1 + y2 + y3 + y4 + y5 = 11 ​ ​ ​ ​ ​

Applying the stars and bars formula:

11 + 5 − 1 15
( )=( )
5−1 4
​ ​

Computing:

15 15! 15 × 14 × 13 × 12
( )= = = 1365
4 4!(11!) 4×3×2×1
​ ​ ​

Thus, the number of solutions is 1,365.

(c) 0 ≤ xi ≤ 10 ​

This is a bounded stars and bars problem, which requires inclusion-exclusion.

1. Total solutions (without bounds on xi ): ​

Using stars and bars:

21 + 5 − 1 25
( )=( )
5−1 4
​ ​

Computing:

25 25! 25 × 24 × 23 × 22
( )= = = 12650
4 4!(21!) 4×3×2×1
​ ​ ​

2. Subtract cases where at least one xi > 10: ​

If one variable (say x1 ) exceeds 10, define x′1 = x1 − 11, where x′1 ≥ 0. The equation becomes:
​ ​ ​ ​

x′1 + x2 + x3 + x4 + x5 = 10 ​ ​ ​ ​ ​

The number of solutions:

10 + 5 − 1 14
( )=( )
5−1 4
​ ​

Computing:

14 14! 14 × 13 × 12 × 11
( )= = = 1001
4 4!(10!) 4×3×2×1
​ ​ ​

Since any of the 5 variables could exceed 10, we subtract:

5 × 1001 = 5005

3. Add cases where at least two xi > 10 (over-subtracted earlier): ​

If two variables exceed 10, say x1 , x2 , define: ​ ​

x′1 = x1 − 11, ​ ​
x′2 = x2 − 11 ​ ​

The equation becomes:

x′1 + x′2 + x3 + x4 + x5 = −1 ​ ​ ​ ​ ​

This has no non-negative integer solutions, so this case contributes 0.

Since cases for three or more variables exceeding 10 also result in negative values, they also
contribute 0.

Thus, the final count is:

12650 − 5005 = 7645

Thus, the number of solutions is 7,645.

Final Answers

(a) 4,845
(b) 1,365
(c) 7,645
Let me know if you need any clarifications! 😊

The problem asks for the number of positive integers less than 1,000,000 (i.e., having at most 6 digits)
where the sum of the digits is 19.

Understanding the Problem

Each number can have at most 6 digits (i.e., d1 d2 d3 d4 d5 d6 ), where each digit di is between 0 and 9, ​ ​ ​ ​ ​ ​ ​

and the total sum of these digits must be 19.

Approach: Stars and Bars

This is a combinatorial problem, specifically a "stars and bars" problem, where we need to find the
number of non-negative integer solutions to:

d1 + d2 + d3 + d4 + d5 + d6 = 19
​ ​ ​ ​ ​ ​

where each di is a digit between 0 and 9. ​

Step 1: Unbounded Stars and Bars

Ignoring the digit limit (0-9), the number of non-negative integer solutions is:

19 + 6 − 1 24
( )=( )
6−1 5
​ ​

Computing:

24 24! 24 × 23 × 22 × 21 × 20
( )= = = 42504
5 5!(19!) 5×4×3×2×1
​ ​ ​

However, this count includes cases where one or more digits exceed 9, which we must remove using
inclusion-exclusion.

Step 2: Removing Invalid Cases

If one digit exceeds 9, define d′1 ​ = d1 − 10 (so d′1 ≥ 0). The equation transforms to:
​ ​

d′1 + d2 + d3 + d4 + d5 + d6 = 9
​ ​ ​ ​ ​ ​

The number of solutions to this is:

9+6−1 14
( )=( )
6−1 5
​ ​

Computing:

14 14! 14 × 13 × 12 × 11 × 10
( )= = = 2002
5 5!(9!) 5×4×3×2×1
​ ​ ​

Since any of the 6 digits could exceed 9, we subtract:

6 × 2002 = 12012

Step 3: Adding Back Over-Subtracted Cases

If two digits exceed 9, define d′1 ​ = d1 − 10 and d′2 = d2 − 10, so the equation becomes:
​ ​ ​

d′1 + d′2 + d3 + d4 + d5 + d6 = −1
​ ​ ​ ​ ​ ​

Since non-negative integer solutions are not possible for a negative sum, this case contributes 0.

Final Count

42504 − 12012 = 30492

Thus, the final answer is:

30, 492

Let me know if you need further clarification! 😊

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