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Question: How do you evaluate \({}^5{C_2} + {}^5{C_1}\)?...

How do you evaluate 5C2+5C1{}^5{C_2} + {}^5{C_1}?

Explanation

Solution

The given question is of Combination. Let us first know what a combination is in context to mathematics. A combination is one of the processes of selection of all or part of a group of things or objects, without regard to the order in which things are being selected. It is expressed in the form of nCr{}^n{C_r}. The number of combinations of nn objects taken rr at a time is expressed as nCr{}^n{C_r}, where
nCr=n!r!(n−r)!=n(n−1)(n−2).....(n−r+1)r(r−1)(r−2)......(2)(1){}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{n(n - 1)(n - 2).....(n - r + 1)}}{{r(r - 1)(r - 2)......(2)(1)}}
We will use the above given formula of nCr{}^n{C_r} to solve the question.

Complete Step by Step Solution:
The given expression is 5C2+5C1{}^5{C_2} + {}^5{C_1}. The formula which we will use to solve the expression is stated as:
nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
We will find the values of 5C2{}^5{C_2} and 5C1{}^5{C_1} separately and then add them up together to get the final answer. Therefore,
For 5C2{}^5{C_2}:
We have n=5n = 5 and r=2r = 2. On substituting these values of nn and rr in the formula, we will get
⇒nCr=n!r!(n−r)!\Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
⇒5C2=5!2!(5−2)!\Rightarrow {}^5{C_2} = \dfrac{{5!}}{{2!(5 - 2)!}}
On further simplifying, we will get
⇒5C2=5!2!3!\Rightarrow {}^5{C_2} = \dfrac{{5!}}{{2!3!}}
It must be known that the factorial of a number nn is expressed as n!=n(n−1)(n−2)(n−3)....(2)(1)n! = n(n - 1)(n - 2)(n - 3)....(2)(1). So from this formula for n!n!, we will find the values of 5!5!, 2!2! and 3!3!, such that
⇒5C2=(5×4×3×2×1)(2×1)(3×2×1)\Rightarrow {}^5{C_2} = \dfrac{{(5 \times 4 \times 3 \times 2 \times 1)}}{{(2 \times 1)(3 \times 2 \times 1)}}
⇒5C2=12012\Rightarrow {}^5{C_2} = \dfrac{{120}}{{12}}
On further simplifying the obtained fraction into its simplest form, we will get
⇒5C2=10\Rightarrow {}^5{C_2} = 10
Hence we have 5C2=10{}^5{C_2} = 10.
Now, for 5C1{}^5{C_1}:
We have n=5n = 5 and r=1r = 1. On substituting these values of nn and rr in the formula, we will get
⇒nCr=n!r!(n−r)!\Rightarrow {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}
⇒5C1=5!2!(5−1)!\Rightarrow {}^5{C_1} = \dfrac{{5!}}{{2!(5 - 1)!}}
On further simplifying, we will get
⇒5C1=5!2!4!\Rightarrow {}^5{C_1} = \dfrac{{5!}}{{2!4!}}
Since for a number nn, n!=n(n−1)(n−2)(n−3)....(2)(1)n! = n(n - 1)(n - 2)(n - 3)....(2)(1). So from this formula for n!n!, we will find the values of 5!5!, 2!2! and 4!4!, such that
⇒5C1=(5×4×3×2×1)(2×1)(4×3×2×1)\Rightarrow {}^5{C_1} = \dfrac{{(5 \times 4 \times 3 \times 2 \times 1)}}{{(2 \times 1)(4 \times 3 \times 2 \times 1)}}
⇒5C1=12048\Rightarrow {}^5{C_1} = \dfrac{{120}}{{48}}
On further simplifying the obtained fraction into its simplest form, we will get
⇒5C1=52\Rightarrow {}^5{C_1} = \dfrac{5}{2}
Hence we have 5C1=52{}^5{C_1} = \dfrac{5}{2}.
Since 5C2=10{}^5{C_2} = 10 and 5C1=52{}^5{C_1} = \dfrac{5}{2}, we will find the value of 5C2+5C1{}^5{C_2} + {}^5{C_1} as
⇒5C2+5C1=10+52\Rightarrow {}^5{C_2} + {}^5{C_1} = 10 + \dfrac{5}{2}
On further simplifying, we will get
⇒5C2+5C1=252\Rightarrow {}^5{C_2} + {}^5{C_1} = \dfrac{{25}}{2}

Hence, on evaluating 5C2+5C1{}^5{C_2} + {}^5{C_1}, we get 252\dfrac{{25}}{2} as the answer.

Note:
It must be known that the factorial of 00, i.e. 0!0! =1 = 1. Therefore the value of nC0{}^n{C_0} can be stated as:
$$$$$ \Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!(n - 0)!}} \Rightarrow {}^n{C_0} = \dfrac{{n!}}{{0!n!}}Sincethe Since then!inboththenumeratoranddenominatordivideseachother,weareleftwithin both the numerator and denominator divides each other, we are left with \Rightarrow {}^n{C_0} = \dfrac{1}{{0!}}Since Since0! = 1,thereforewehave, therefore we have \Rightarrow {}^n{C_0} = 1$