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This is one of our many quizzes on Combinations and Pascal's triangle. Click on this to access more such competitive quizzes on Algebra.

Combinations and Pascal's triangle Quiz 2

mathematics

  A student is trying to calculate the number of permutations by looking at the pascal’s triangle. If the $5\text{th}$ row and $1\text{st}$ item is given to the student, then calculate the number permutations. <br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/553e1c7e-1ca5-431c-8d0c-fbe24e06bf733083635976275538689.png"/>

 From $C\left( n,r \right)$, $n$ represents the row in pascal’s triangle whilst $r$ represents the item in pascal’s triangle. Identify the combination and multiply it by $r!$ to calculate the number of permutations.

$\text{A combination in the pascal’s triangle is denoted by the row which represents the sample or n }\\ \text{The item denotes the object or r in }C\left( n,r \right)\\ \text{Therefore, }C\left( n,r \right)=5\text{ is represented by }\text{5th}\text{ row and }\text{1st}\text{ item in pascals triangle. }\\ \text{Since }C\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}$ and $P\left( n,r \right)=\dfrac{n!}{\left( n-r \right)!}\\  P\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}\times r! \\ \Rightarrow P\left( 5,1 \right)=\dfrac{5!}{1!\left( 5-1 \right)!}\times 1!=\dfrac{5!}{\left( 5-1 \right)!} \\  \Rightarrow P\left( 5,1 \right)=\dfrac{5!}{4!}=5$

   If $n-r=3$ where the binomial expression ${{\left( a+b \right)}^{n}}$ has $7$ zeroes, Calculate the value of $C\left( n,r \right)$ and the product $nr$ satisfying the equation $n+r=11$. 

 Calculate the degree of the polynomial using the number of zeroes of the function and thus, calculate the value of $r$. Using the values of $n$ and $r$, evaluate $C\left( n,r \right)$ and the product of $n$ and $r$. Check if the identified values of $n$ and $r$ satisfy the equation $n+r=11$. .

 $ \text{If }{{\left( a+b \right)}^{n}}\text{ has 7 zeros, that suggests that the degree of the polynomial is 7} \text{Hence, n=7 }\\ 7-r=3 \\ \Rightarrow r=4 \\ C\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!} \\ \Rightarrow C\left( 7,4 \right)=\dfrac{7!}{4!\left( 7-4 \right)!} \\ \Rightarrow C\left( 7,4 \right)=\dfrac{7!}{4!3!} \\ \Rightarrow C\left( 7,4 \right)=\dfrac{210}{6}=35 \\ nr=7\times 4 \\ \Rightarrow nr=28 \\ n+r=11 \\ \Rightarrow 7+4=11 \\ \text{Thus, the values of n and r satisfy the equation n+r=11}$

 A binomial expression ${{\left( a+b \right)}^{n}}$ has $5$ zeroes. If one of the coefficients of the binomial expression is found out to be $10$, then calculate the value of $r$ where $r$ represents the number of objects from a sample $n$ in $C\left( n,r \right)$. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/05371720-b83d-4239-a43c-719ed485f63a1403325877214698855.png"/>

 Calculate the degree of the polynomial using the number of zeroes of the function and thus, calculate the value of $n$. Use the pascal’s triangle to identify which item has the coefficient $10$ in the row represented by the value of $n$. 

 $ \text{If }{{\left( a+b \right)}^{n}}\text{ has 5 zeros, that suggests that the degree of the polynomial is 5.}\\ \text{Hence, n=5}\\ \text{From pascal’s triangle, the }5\text{th}\text{ row has two coefficients that have a value of 10 }\\ \text{Item 2 and 3 have a value of 10}\\ \text{Since the item in pascal’s triangle denotes the object or r in }C\left( n,r \right)\\ \text{The values of r are 2 and 3}$

  It is given that two combinations $C\left( n-1,r-1 \right)=5$ with $C\left( n-1,r \right)=10$, calculate the value of the $\dfrac{C\left( n,r \right)}{\left( n+r \right)\left( n-r \right)}$

 Express $C\left( n,r \right)$ as the sum of $C\left( n-1,r-1 \right)$ and $C\left( n-1,r \right)$. Using the value of $C\left( n,r \right)$, evaluate the value of $n$ and $r$ to calculate $\left( n+r \right)\left( n-r \right)$. Use the aforementioned to find the value of $\dfrac{C\left( n,r \right)}{\left( n+r \right)\left( n-r \right)}$.

 $ \text{The pascal’s triangle has values which are the sum of the above two values. }\\ \text{Since the row represented by the value of n}\\ \text{And the item represented by the value of r form }C\left( n,r \right)\text{ in pascal’s triangle,}\\ C\left( n,r \right)\text{ can be expressed as the sum of }C\left( n-1,r-1 \right)\text{ and }C\left( n-1,r \right)\\ C\left( n-1,r-1 \right)=5\\ C\left( n-1,r \right)=10\\ \text{Thus, }C\left( n,r \right)=C\left( n-1,r-1 \right)+C\left( n-1,r \right)=15 \\ C\left( n-1,r \right)=\dfrac{\left( n-1 \right)!}{r!\left( n-1-r \right)!}=5, C\left( n,r \right)=\dfrac{n!}{r!\left( n-r \right)!}=15\\ \text{Using pascal’s triangle, n=6,r=2}\\ \text{Hence, }\left( n+r \right)\left( n-r \right)=32\\ \text{Therefore, }\dfrac{C\left( n,r \right)}{\left( n+r \right)\left( n-r \right)}=\dfrac{15}{32}$

 Identify a binomial expression that satisfies the combination $C\left( n,r \right)$ represented by the $\text{3rd}$ row and the $\text{1st}$ item from the pascal’s triangle. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/349b294b-e1ed-4e25-975c-190b6a9acea82592923562924011516.png"/>

 Express the formula of binomial expansion as a summation. Use the row and the item in pascal’s triangle to calculate the value of the combination and thus, evaluate the sum to identify the binomial expression.

 $\text{The binomial theorem can be expressed as a summation}\\ \text{For a binomial expression }{{\left( a+b \right)}^{n}}, {{\left( a+b \right)}^{n}}=\sum\limits_{r=0}^{n}{C\left( n,r \right){{x}^{n-r}}{{y}^{r}}}\\ \text{A combination in the pascal’s triangle is denoted by the row which represents the sample or n}\\ \text{The item denotes the object or r in }C\left( n,r \right)\\ \text{Therefore, }C\left( n,r \right)=3\text{ is represented by }\text{3rd}\text{ row and }\text{1st}\text{ item in pascals triangle. }\text{Using the value of n and r,}\\ {{\left( a+b \right)}^{3}}=\sum\limits_{r=0}^{3}{C\left( 3,r \right){{x}^{3-r}}{{y}^{r}}}\\ \text{For r=0, }C\left( 3,0 \right)=1 \\ \text{For r=1, }C\left( 3,1 \right)=3\\ \text{For r=2, }C\left( 3,2 \right)=3 \\ \text{For r=3 }, C\left( 3,3 \right)=1\\ \text{Evaluating the sum, }\sum\limits_{r=0}^{3}{C\left( 3,r \right){{x}^{3-r}}{{y}^{r}}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$