Write the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\].
Answer
Verified559.8k+ views
Hint: 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 \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
Complete step by step solution:
We have to find the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\]
Here the elements of the sum are \[^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}\]
Use the formula of combinations i.e. \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] to find each of the elements of the sum.
We substitute the value of \[n = 5,r = 1\] in the formula of combination.
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{(5 - 1)!1!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{4!1!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5 \times 4!}}{{4!1!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_1} = \dfrac{5}{{1!}}\]
Substitute the value of \[1! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_1} = 5\]..................… (1)
We substitute the value of \[n = 5,r = 2\] in the formula of combination.
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{(5 - 2)!2!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{3!2!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4 \times 3!}}{{3!2!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{{2!}}\]
Substitute the value of \[2! = 2 \times 1\] in the denominator
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{2}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_2} = 10\]....................… (2)
We substitute the value of \[n = 5,r = 3\] in the formula of combination.
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{(5 - 3)!3!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{2!3!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4 \times 3!}}{{2!3!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{{2!}}\]
Substitute the value of \[2! = 2 \times 1\] in the denominator
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{2}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_3} = 10\].................… (3)
We substitute the value of \[n = 5,r = 1\] in the formula of combination.
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{(5 - 4)!4!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{1!4!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5 \times 4!}}{{1!4!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_4} = \dfrac{5}{{1!}}\]
Substitute the value of \[1! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_4} = 5\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_4} = 5\]..............… (4)
We substitute the value of \[n = 5,r = 5\] in the formula of combination.
\[{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{(5 - 5)!5!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{0!5!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_5} = \dfrac{1}{{0!}}\]
Substitute the value of \[0! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_5} = 1\]...............… (5)
We substitute the values of \[^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}\]in the sum\[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 5 + 10 + 10 + 5 + 1\]
Add the terms in RHS of the equation
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31\]
\[\therefore \]The value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] is 31.
Note: Students are likely to make the mistake of substituting the value of \[0! = 0\] which is wrong. Keep in mind this value of factorial of zero is fixed as 1, similarly the value of \[1! = 1\] is fixed.
Alternate method:
We can solve for the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] using the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
Here the value of \[n = 5\]
Put\[n = 5\]; \[r = 1\] in the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
\[{ \Rightarrow ^5}{C_1}{ = ^5}{C_{5 - 1}}\]
\[{ \Rightarrow ^5}{C_1}{ = ^5}{C_4}\]................… (1)
Put\[n = 5\]; \[r = 2\] in the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
\[{ \Rightarrow ^5}{C_2}{ = ^5}{C_{5 - 2}}\]
\[{ \Rightarrow ^5}{C_2}{ = ^5}{C_3}\].................… (2)
Substitute the values from equations (1) and (2) in the equation given in the question
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_2}{ + ^5}{C_1}{ + ^5}{C_5}} \right)\]
Add like terms
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2{ \times ^5}{C_1} + 2{ \times ^5}{C_2}{ + ^5}{C_5}} \right)\]
We know formula of combination is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\] .
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{(5 - 1)!1!}} + 2 \times \dfrac{{5!}}{{(5 - 2)!2!}} + \dfrac{{5!}}{{(5 - 5)!5!}}} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{4!1!}} + 2 \times \dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)\]
Write numerator using factorial formula
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5 \times 4!}}{{4!1!}} + 2 \times \dfrac{{5 \times 4 \times 3!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{5}{{1!}} + 2 \times \dfrac{{5 \times 4}}{{2!}} + \dfrac{1}{{0!}}} \right)\]
Put \[1! = 1,2! = 2,0! = 1\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times 5 + 5 \times 4 + 1} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {10 + 20 + 1} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31\]
\[\therefore \]The value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] is 31.
* Formula of combination is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\]
Complete step by step solution:
We have to find the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\]
Here the elements of the sum are \[^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}\]
Use the formula of combinations i.e. \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\] to find each of the elements of the sum.
We substitute the value of \[n = 5,r = 1\] in the formula of combination.
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{(5 - 1)!1!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5!}}{{4!1!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_1} = \dfrac{{5 \times 4!}}{{4!1!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_1} = \dfrac{5}{{1!}}\]
Substitute the value of \[1! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_1} = 5\]..................… (1)
We substitute the value of \[n = 5,r = 2\] in the formula of combination.
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{(5 - 2)!2!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5!}}{{3!2!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4 \times 3!}}{{3!2!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{{2!}}\]
Substitute the value of \[2! = 2 \times 1\] in the denominator
\[{ \Rightarrow ^5}{C_2} = \dfrac{{5 \times 4}}{2}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_2} = 10\]....................… (2)
We substitute the value of \[n = 5,r = 3\] in the formula of combination.
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{(5 - 3)!3!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5!}}{{2!3!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4 \times 3!}}{{2!3!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{{2!}}\]
Substitute the value of \[2! = 2 \times 1\] in the denominator
\[{ \Rightarrow ^5}{C_3} = \dfrac{{5 \times 4}}{2}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_3} = 10\].................… (3)
We substitute the value of \[n = 5,r = 1\] in the formula of combination.
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{(5 - 4)!4!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5!}}{{1!4!}}\]
Write the numerator using method of factorial i.e. \[n! = n \times (n - 1)!\]
\[{ \Rightarrow ^5}{C_4} = \dfrac{{5 \times 4!}}{{1!4!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_4} = \dfrac{5}{{1!}}\]
Substitute the value of \[1! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_4} = 5\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_4} = 5\]..............… (4)
We substitute the value of \[n = 5,r = 5\] in the formula of combination.
\[{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{(5 - 5)!5!}}\]
Calculate the difference in the denominator
\[{ \Rightarrow ^5}{C_5} = \dfrac{{5!}}{{0!5!}}\]
Cancel the same terms from numerator and denominator.
\[{ \Rightarrow ^5}{C_5} = \dfrac{1}{{0!}}\]
Substitute the value of \[0! = 1\] in the denominator
\[{ \Rightarrow ^5}{C_5} = 1\]...............… (5)
We substitute the values of \[^5{C_1}{,^5}{C_2}{,^5}{C_3}{,^5}{C_4}{,^5}{C_5}\]in the sum\[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 5 + 10 + 10 + 5 + 1\]
Add the terms in RHS of the equation
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31\]
\[\therefore \]The value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] is 31.
Note: Students are likely to make the mistake of substituting the value of \[0! = 0\] which is wrong. Keep in mind this value of factorial of zero is fixed as 1, similarly the value of \[1! = 1\] is fixed.
Alternate method:
We can solve for the value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] using the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
Here the value of \[n = 5\]
Put\[n = 5\]; \[r = 1\] in the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
\[{ \Rightarrow ^5}{C_1}{ = ^5}{C_{5 - 1}}\]
\[{ \Rightarrow ^5}{C_1}{ = ^5}{C_4}\]................… (1)
Put\[n = 5\]; \[r = 2\] in the formula \[^n{C_r}{ = ^n}{C_{n - r}}\]
\[{ \Rightarrow ^5}{C_2}{ = ^5}{C_{5 - 2}}\]
\[{ \Rightarrow ^5}{C_2}{ = ^5}{C_3}\].................… (2)
Substitute the values from equations (1) and (2) in the equation given in the question
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_2}{ + ^5}{C_1}{ + ^5}{C_5}} \right)\]
Add like terms
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2{ \times ^5}{C_1} + 2{ \times ^5}{C_2}{ + ^5}{C_5}} \right)\]
We know formula of combination is given by \[^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}\], where factorial is expanded by the formula \[n! = n \times (n - 1)! = n \times (n - 1) \times (n - 2)!.... = n \times (n - 1) \times (n - 2)....3 \times 2 \times 1\] .
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{(5 - 1)!1!}} + 2 \times \dfrac{{5!}}{{(5 - 2)!2!}} + \dfrac{{5!}}{{(5 - 5)!5!}}} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5!}}{{4!1!}} + 2 \times \dfrac{{5!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)\]
Write numerator using factorial formula
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{{5 \times 4!}}{{4!1!}} + 2 \times \dfrac{{5 \times 4 \times 3!}}{{3!2!}} + \dfrac{{5!}}{{0!5!}}} \right)\]
Cancel the same terms from numerator and denominator.
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times \dfrac{5}{{1!}} + 2 \times \dfrac{{5 \times 4}}{{2!}} + \dfrac{1}{{0!}}} \right)\]
Put \[1! = 1,2! = 2,0! = 1\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {2 \times 5 + 5 \times 4 + 1} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = \left( {10 + 20 + 1} \right)\]
\[ \Rightarrow \left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right) = 31\]
\[\therefore \]The value of \[\left( {^5{C_1}{ + ^5}{C_2}{ + ^5}{C_3}{ + ^5}{C_4}{ + ^5}{C_5}} \right)\] is 31.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success
Master Class 11 Computer Science: Engaging Questions & Answers for Success
Master Class 11 Maths: Engaging Questions & Answers for Success
Master Class 11 Chemistry: Engaging Questions & Answers for Success
Master Class 11 Economics: Engaging Questions & Answers for Success
Master Class 11 Accountancy: Engaging Questions & Answers for Success
Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
What are Quantum numbers Explain the quantum number class 11 chemistry CBSE