Question

hello my dear brothers and sisters please help me in this
with step by step explanation wrong answer will be reported

this is class 8 math IIT jee level 4 ​

Answer 1

Hello sis

23. x+y = 1

(x+y)ⁿ = ⁿC₀ xⁿ y⁰ + ⁿC₁ xⁿ ⁻¹ y + ...... + ⁿCₙ x⁰ yⁿ

When you substitute 0, the whole term will become zero

As you know that x+y = y+x,

y+x = 1

So, when we write (y+x)ⁿ :

(y+x)ⁿ = ⁿC₀ yⁿ x⁰ + ⁿC₁ yⁿ⁻¹ x +....+ ⁿCₙ xⁿ y⁰

Now differentiate on both sides:

n(x+y)ⁿ⁻¹ = ⁿC₁ yⁿ⁻¹ + 2 ⁿC₂ x yⁿ⁻² +........+ n ⁿCₙ xⁿ⁻¹  

multiplying by x on both sides, you will get:

nx (y+x)ⁿ⁻¹  = ⁿC₁ x yⁿ⁻¹ + 2 C₂ x² yⁿ⁻² +.....+ ⁿCₙ xⁿ⁻¹ × x

= C₁ x yⁿ⁻¹ + 2 C₂ x² yⁿ⁻² +.....+ ⁿCₙ xⁿ

So, you need the value of the expression that is at the left side

substitute y+x = 1

You will get:

nx (1)ⁿ⁻¹ = C₁ x yⁿ⁻¹ + 2 C₂ x² yⁿ⁻² +.....+ ⁿCₙ xⁿ

Anything power 1 =1

So, nx = C₁ x yⁿ⁻¹ + 2 C₂ x² yⁿ⁻² +.....+ ⁿCₙ xⁿ

So, your answer is option 3

24. ³⁴C₅​+ $\Sigma^4_r_=_0$  $^3^8^-^xC_4$​

= ³⁴C₅+ ³⁸C₄+ ³⁷C₄​+ ³⁶C₄​ + ³⁵C₄​+ ³⁴C₄​

=(³⁴C₅​+³⁴C₄) + ³⁵C₄​+ ³⁶ C₄​+ ³⁷C₄​+ ³⁸C₄​

=³⁵C₅ + ³⁵C₄ + ³⁶C₄ + ³⁷C₄ + ³⁸C₄   ​(∵ ⁿC$_r$​+ ⁿC$_r_-_1$= ⁿ⁺¹C$_r$)

=³⁶C₅ + ³⁶C₄ ​+ ³⁷C₄ ​+ ³⁸C₄

= ³⁷C₅ + ³⁷C₅ ​+ ³⁸C₄

= ³⁸C₅ ​+ ³⁸C₄

= ³⁹C₅

=34! × 5! × 39!​

=5 × 4 × 3 × 2 × 1 × 39 × 38 × 37 × 36 × 35    ​(∵ ⁿC$_r$​=r!(n−r)!n!​)

Hence the value is ³⁹C₅ which is option 1

25. ⁿP​$_r$ = (n−r)! n!​

Now 5040 =10(9)(8)(7)

= $\frac{10!}{6!}$

= $\frac{10!}{(10-4)!}$

= ¹⁰P₄

Hence, n= 10 and r= 4

Option 2 is your answer

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