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What is the number of distinct terms in the expansion of (a+b+c)20 ? [CAT, 2008]

 

a. 231 

b. 253 

c. 242        

d. 210

e. 228

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Solution:

What is the number of distinct terms in the expansion

Before we solve this question, let’s solve similar questions to understand it clearly because the like may be given in the next turn.

Example 1

(x+y+z)30

Using the formula below, we can solve this type of question easily.

\frac { (p)+(k-1)! }{ (p!) (k-1)! }

Where P is the power and k is the number of terms. 

In example 1, P=30 and k=3

\frac { (30)+ (3-1)! }{ (30!) (3-1)! }

\frac { 30 +2! }{ 30!\quad 2! }

 \frac { 32! }{ 30!\quad \times \quad 2\quad \times \quad 1 }

\frac { 32\quad \times \quad 31\quad \times \quad 30! }{ 30!\quad \times \quad 2\quad \times \quad 1 }

\frac { 32\times 31 }{ 2\times 1 }

\frac { 16\times 31 }{ 1 }

16×31= 496

Sample Problem 2

 (a+b+c+d)40

 

In example 1, P=40 and k=4

\frac { (40)+\quad (4-1)! }{ (40!) (4-1)! }

\frac { 40\quad +\quad 3! }{ 40!\quad \quad 3! }

\frac { 43! }{ 40!\times \quad 3\quad \times \quad 2\quad \times \quad 1 }

\frac { 43\quad \times \quad 42\quad \times \quad 41\quad \times \quad 40! }{ 40!\quad \times \quad 3\quad \times \quad 2\quad \times \quad 1 }

\frac { 43\times 42\times 41 }{ \quad \quad 3\times 2\times 1 }

\frac { 43\times 42\times 41 } { \quad \quad 6 }

43 x 7 x 41 = 287 x 43 = 12341

Now the given question, CAT, 2008

(a+b+c)20

Where P is the power and k is the number of terms.

P=20 and k=3

\frac { (20)+\quad (3-1)! }{ (20!) (3-1)! }

\frac { 20\quad +\quad 2! }{ 20!\quad \quad 2! }

\frac { 22! }{ 20!\quad \times \quad 2\quad \times \quad 1 }

\frac { 22\quad \times \quad 21\quad \times \quad 20! }{ 20!\quad \times \quad 2\quad \times \quad 1 }

\frac { 22\quad \times \quad 21\quad }{ 1 }

\frac { 11\quad \times \quad 21 }{ 1 }

11 x 21=231

Hence, option a is correct

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