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Question: Write the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\]....

Write the value of (5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right).

Explanation

Solution

We use the formula of combination for each of the elements in the sum. Substitute the corresponding values of ‘n’ and ‘r’ in the formula of combination for respective terms.

  • Formula of combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}, where factorial is expanded by the formula n!=n×(n−1)!=n×(n−1)×(n−2)!....=n×(n−1)×(n−2)....3×2×1n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1

Complete step by step solution:
We have to find the value of (5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)
Here the elements of the sum are 5C1,5C2,5C3,5C4,5C5^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}
Use the formula of combinations i.e. nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} to find each of the elements of the sum.
We substitute the value of n=5,r=1n = 5,r = 1 in the formula of combination.
⇒5C1=5!(5−1)!1!{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{(5 - 1)!1!}}
Calculate the difference in the denominator
⇒5C1=5!4!1!{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{4!1!}}
Write the numerator using method of factorial i.e. n!=n×(n−1)!n! = n \times (n - 1)!
⇒5C1=5×4!4!1!{ \Rightarrow ^5}{C_1} = \dfrac{{5 \times 4!}}{{4!1!}}
Cancel the same terms from numerator and denominator.
⇒5C1=51!{ \Rightarrow ^5}{C_1} = \dfrac{5}{{1!}}
Substitute the value of 1!=11! = 1 in the denominator
⇒5C1=5{ \Rightarrow ^5}{C_1} = 5..................… (1)
We substitute the value of n=5,r=2n = 5,r = 2 in the formula of combination.
⇒5C2=5!(5−2)!2!{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{(5 - 2)!2!}}
Calculate the difference in the denominator
⇒5C2=5!3!2!{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{3!2!}}
Write the numerator using method of factorial i.e. n!=n×(n−1)!n! = n \times (n - 1)!
⇒5C2=5×4×3!3!2!{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4 \times 3!}}{{3!2!}}
Cancel the same terms from numerator and denominator.
⇒5C2=5×42!{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{{2!}}
Substitute the value of 2!=2×12! = 2 \times 1 in the denominator
⇒5C2=5×42{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{2}
Cancel the same terms from numerator and denominator.
⇒5C2=10{ \Rightarrow ^5}{C_2} = 10....................… (2)
We substitute the value of n=5,r=3n = 5,r = 3 in the formula of combination.
⇒5C3=5!(5−3)!3!{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{(5 - 3)!3!}}
Calculate the difference in the denominator
⇒5C3=5!2!3!{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{2!3!}}
Write the numerator using method of factorial i.e. n!=n×(n−1)!n! = n \times (n - 1)!
⇒5C3=5×4×3!2!3!{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4 \times 3!}}{{2!3!}}
Cancel the same terms from numerator and denominator.
⇒5C3=5×42!{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{{2!}}
Substitute the value of 2!=2×12! = 2 \times 1 in the denominator
⇒5C3=5×42{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{2}
Cancel the same terms from numerator and denominator.
⇒5C3=10{ \Rightarrow ^5}{C_3} = 10.................… (3)
We substitute the value of n=5,r=1n = 5,r = 1 in the formula of combination.
⇒5C4=5!(5−4)!4!{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{(5 - 4)!4!}}
Calculate the difference in the denominator
⇒5C4=5!1!4!{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{1!4!}}
Write the numerator using method of factorial i.e. n!=n×(n−1)!n! = n \times (n - 1)!
⇒5C4=5×4!1!4!{ \Rightarrow ^5}{C_4} = \dfrac{{5 \times 4!}}{{1!4!}}
Cancel the same terms from numerator and denominator.
⇒5C4=51!{ \Rightarrow ^5}{C_4} = \dfrac{5}{{1!}}
Substitute the value of 1!=11! = 1 in the denominator
⇒5C4=5{ \Rightarrow ^5}{C_4} = 5
Cancel the same terms from numerator and denominator.
⇒5C4=5{ \Rightarrow ^5}{C_4} = 5..............… (4)
We substitute the value of n=5,r=5n = 5,r = 5 in the formula of combination.
⇒5C5=5!(5−5)!5!{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{(5 - 5)!5!}}
Calculate the difference in the denominator
⇒5C5=5!0!5!{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{0!5!}}
Cancel the same terms from numerator and denominator.
⇒5C5=10!{ \Rightarrow ^5}{C_5} = \dfrac{1}{{0!}}
Substitute the value of 0!=10! = 1 in the denominator
⇒5C5=1{ \Rightarrow ^5}{C_5} = 1...............… (5)
We substitute the values of 5C1,5C2,5C3,5C4,5C5^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}in the sum(5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)
⇒(5C1+5C2+5C3+5C4+5C5)=5+10+10+5+1\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 5 + 10 + 10 + 5 + 1
Add the terms in RHS of the equation
⇒(5C1+5C2+5C3+5C4+5C5)=31\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31

∴\therefore The value of (5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) is 31.

Note: Students are likely to make the mistake of substituting the value of 0!=00! = 0 which is wrong. Keep in mind this value of factorial of zero is fixed as 1, similarly the value of 1!=11! = 1 is fixed.
Alternate method:
We can solve for the value of (5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) using the formula nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}}
Here the value of n=5n = 5
Putn=5n = 5; r=1r = 1 in the formula nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}}
⇒5C1=5C5−1{ \Rightarrow ^5}{C_1}{ = ^5}{C_{5 - 1}}
⇒5C1=5C4{ \Rightarrow ^5}{C_1}{ = ^5}{C_4}................… (1)
Putn=5n = 5; r=2r = 2 in the formula nCr=nCn−r^n{C_r}{ = ^n}{C_{n - r}}
⇒5C2=5C5−2{ \Rightarrow ^5}{C_2}{ = ^5}{C_{5 - 2}}
⇒5C2=5C3{ \Rightarrow ^5}{C_2}{ = ^5}{C_3}.................… (2)
Substitute the values from equations (1) and (2) in the equation given in the question
⇒(5C1+5C2+5C3+5C4+5C5)=(5C1+5C2+5C2+5C1+5C5)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_2}{ + ^5}{C_1}{ + ^5}{C_5}} \right)
Add like terms
⇒(5C1+5C2+5C3+5C4+5C5)=(2×5C1+2×5C2+5C5)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2{ \times ^5}{C_1} + 2{ \times ^5}{C_2}{ + ^5}{C_5}} \right)
We know formula of combination is given by nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}, where factorial is expanded by the formula n!=n×(n−1)!=n×(n−1)×(n−2)!....=n×(n−1)×(n−2)....3×2×1n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1 .
⇒(5C1+5C2+5C3+5C4+5C5)=(2×5!(5−1)!1!+2×5!(5−2)!2!+5!(5−5)!5!)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{(5 - 1)!1!}} + 2 \times \dfrac{{5!}}{{(5 - 2)!2!}} + \dfrac{{5!}}{{(5 - 5)!5!}}} \right)
⇒(5C1+5C2+5C3+5C4+5C5)=(2×5!4!1!+2×5!3!2!+5!0!5!)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{4!1!}} + 2 \times \dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)
Write numerator using factorial formula
⇒(5C1+5C2+5C3+5C4+5C5)=(2×5×4!4!1!+2×5×4×3!3!2!+5!0!5!)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5 \times 4!}}{{4!1!}} + 2 \times \dfrac{{5 \times 4 \times 3!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)
Cancel the same terms from numerator and denominator.
⇒(5C1+5C2+5C3+5C4+5C5)=(2×51!+2×5×42!+10!)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{5}{{1!}} + 2 \times \dfrac{{5 \times 4}}{{2!}} + \dfrac{1}{{0!}}} \right)
Put 1!=1,2!=2,0!=11! = 1,2! = 2,0! = 1
⇒(5C1+5C2+5C3+5C4+5C5)=(2×5+5×4+1)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times 5 + 5 \times 4 + 1} \right)
⇒(5C1+5C2+5C3+5C4+5C5)=(10+20+1)\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {10 + 20 + 1} \right)
⇒(5C1+5C2+5C3+5C4+5C5)=31\Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31
∴\therefore The value of (5C1+5C2+5C3+5C4+5C5)\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) is 31.