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The combinatorial coefficients ""^(n – 1...

The combinatorial coefficients `""^(n – 1)C_(p)` denotes

A

The number of ways in which n things of which p are alike and rest different can be arranged in a circle

B

The number of ways in which p different things can be selected out of n different thing if a particular thing is always excluded

C

Number of ways in which n alike balls can be distributed in p different boxes so that no box remains empty and each box can hold any number of balls

D

The number of ways in which (n – 2) white balls and p black balls can be arranged in a line if no two black balls are together, balls are all alike except for the colour

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To solve the problem regarding the combinatorial coefficient \( C(n-1, p) \), we will analyze the meaning of this expression step by step. ### Step-by-Step Solution: 1. **Understanding the Notation**: The notation \( C(n-1, p) \) or \( \binom{n-1}{p} \) represents the number of ways to choose \( p \) elements from a set of \( n-1 \) elements. This is known as a binomial coefficient. **Hint**: Remember that \( C(n, r) \) is defined as \( \frac{n!}{r!(n-r)!} \). 2. **Identifying the Context**: The problem asks us to interpret what \( C(n-1, p) \) denotes in a combinatorial context. This can relate to various scenarios such as arranging objects, selecting items, or distributing indistinguishable objects into distinguishable boxes. **Hint**: Think about scenarios where you have a total of \( n-1 \) items and you need to select \( p \) from them. 3. **Analyzing Possible Conditions**: - **Condition 1**: The number of ways to arrange \( n \) items where \( p \) are alike and the rest are different. This can often lead to a circular arrangement, which is more complex and may not directly relate to \( C(n-1, p) \). **Hint**: Consider how circular arrangements differ from linear arrangements. - **Condition 2**: The number of ways to choose \( p \) different items from \( n-1 \) items, which directly corresponds to \( C(n-1, p) \). **Hint**: This condition directly matches the definition of the binomial coefficient. - **Condition 3**: The distribution of \( n \) alike objects into \( p \) boxes where no box is empty. This scenario often uses the "stars and bars" theorem, which may not directly relate to \( C(n-1, p) \). **Hint**: Recall the "stars and bars" theorem for distributing indistinguishable objects. - **Condition 4**: Arranging \( n-2 \) alike balls and \( p \) black balls in a line with the restriction that no two black balls are adjacent. This is a combinatorial arrangement problem and may not directly yield \( C(n-1, p) \). **Hint**: Think about how restrictions affect arrangements. 4. **Conclusion**: After analyzing the conditions, we find that the second condition, which states the number of ways to choose \( p \) different items from \( n-1 \) items, directly corresponds to the combinatorial coefficient \( C(n-1, p) \). **Final Answer**: The combinatorial coefficient \( C(n-1, p) \) denotes the number of ways to choose \( p \) items from \( n-1 \) items.
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If C_(0),C_(1),C_(2)…….,C_(n) are the combinatorial coefficient in the expansion of (1+x)^n, n, ne N , then prove that following C_(1)+2C_(2)+3C_(3)+..+n.C_(n)=n.2^(n-1) C_(0)+2C_(1)+3C_(2)+......+(n+1)C_(n)=(n+2)C_(n)=(n+2)2^(n-1) C_(0),+3C_(1)+5C_(2)+.....+(2n+1)C_n =(n+1)2^n (C_0+C_1)(C_1+C_2)(C_2+C_3)......(C_(n-1)+C_n)=(C_0.C_1.C_2....C_(n-1)(n+1)^n)/(n!) 1.C_0^2+3.C_1^2+....+ (2n+1)C_n^2=((n+1)(2n)!)/(n! n!)

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If C_(0), C_(1), C_(2), …, C_(n) denote the binomial coefficients in the expansion of (1 + x)^(n) , then sum_(r=0)^(n)sum_(s=0)^(n)C_(r)C_(s) =

If C_(0), C_(1), C_(2),…, C_(n) denote the binomial coefficients in the expansion of (1 + x)^(n) , then sum_(r=0)^(n)sum_(s=0)^(n)(C_(r) +C_(s))

If C_0,C_1,C_2,.......,C_n denote the binomial coefficients in the expansion of (1+1)^n, then sum_(0ler) sum_(< s le n) (C_r+C_s)

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