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Let  F = ( x + y + z )⁵

Let w = y+z

First term of above expression is only one term

Second term of above expression has w = (y+z) , hence numvber of terms in actual expansion is 2

Third term of above expression has w = (y+z)² , hence numvber of terms in actual expansion is 3

Fourth term of above expression has w = (y+z)^3 , hence numvber of terms in actual expansion is 4

Fifth term of above expression has w = (y+z)^4 , hence numvber of terms in actual expansion is 5

Sixth term of above expression has w = (y+z)^5 , hence numvber of terms in actual expansion is 6

hence number of terms = 1+2+3+4+5+6 = 21

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if n = 5 , then ⁽ⁿ⁺²⁾C₂ = ⁷C₂ = 21

⁷C₂ is number of ways choosing 2 items randomly from 7 items .

Number of combinations of choosing 2 out of 7 is nothing to do to decide the number of terms of ( x+ y +z )⁵

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Hence Answer ( A ) is correct , bur reason (R) is wrong