Question

Prove that 1 + 3C, + 4c2 = 5C3​

Answer 1

Step-by-step explanation:

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Answer 2

1 + ³C₁ + ⁴C₂ = ⁵C₃ is proved

Given :

The expression 1 + ³C₁ + ⁴C₂ = ⁵C₃

To find :

To prove the expression

Solution :

Step 1 of 3 :

Simplify LHS

LHS

$\displaystyle \sf{ = 1 + {}^{3}C_1 + {}^{4}C_2 }$

$\displaystyle \sf{ =1 + \frac{3!}{1!(3 - 1)!} + \frac{4!}{2!(4 - 2)!} }$

$\displaystyle \sf{ =1 + \frac{3!}{1! \: 2!} + \frac{4!}{2! \: 2!} }$

$\displaystyle \sf{ =1 + \frac{3 \times 2!}{1! \: 2!} + \frac{4 \times 3 \times 2!}{2! \: 2!} }$

$\displaystyle \sf{ =1 + 3 + \frac{12}{2} }$

$\displaystyle \sf{ =1 + 3 + 6 }$

$\displaystyle \sf{ =1 0}$

Step 2 of 3 :

Simplify the RHS

RHS

$\displaystyle \sf{ = {}^{5}C_3}$

$\displaystyle \sf{ } = \frac{5!}{3! \: (5 - 3)!}$

$\displaystyle \sf{ } = \frac{5!}{3! \:2!}$

$\displaystyle \sf{ } = \frac{5 \times 4 \times 3!}{3! \:2!}$

$\displaystyle \sf{ } = \frac{5 \times 4 }{2!}$

$\displaystyle \sf{ } = \frac{5 \times 4 }{2}$

$\displaystyle \sf{ } = 10$

Step 3 of 3 :

Prove the expression

We see that LHS = RHS

Hence the proof follows