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Question:

What is the number of distinct terms in the expansion of (a + b + c)²⁰?

Explanation:

Consider (a + b + c)²⁰

Each term in the expansion of (a + b + c)²⁰ can be written as ²⁰C_r × a^x × b^y × c^z, where (x + y + z) = 20

Number of distinct terms in the expansion will be same as the number of distinct pairs of x, y and z.

∴ We have to divide 20 into three parts which can be done by using the distribution rule = ^n+r-1C_r-1

Where,

n is number of things to be distributed.

r is number of parts into which the things are to be distributed.

∴ To divide 20 into 3 parts we have,

^20+3-1C_3-1 = ²²C₂ = $\frac{22\times 21}{2\times 1}$ = 11 × 21 = 231

Alternatively,
This can be solved without using much knowledge of permutations and combinations as follows,

(a + b + c)¹ = a + b + c [i.e. 3 terms = (1 + 2) terms]

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac [i.e. 6 terms = (1 + 2 + 3) terms]

(a + b + c)³ = a³ + b³ + c³ + 6abc + 3ab² + 3ac² + 3a²b + 3bc² + 3a²c + 3b²c [i.e. 10 terms = (1 + 2 + 3 + 4) terms]

Similarly,

(a + b + c)ⁿ will have (1 + 2 + 3 + … + (n + 1)) terms

∴ (a + b + c)²⁰ will have (1 + 2 + 3 + … + 21) = 231 terms

Hence, option (a).