The least positive integer n for which \[^{n-1}{{C}_{5}}{{+}^{n-1}}{{C}_{6}}\text{ }<{{\text{ }}^{n}}{{C}_{7}}\] is?
1. \[14\]
2. \[15\]
3. \[16\]
4. \[28\]
Answer
574.2k+ views
Hint: We can solve this question by using the logic and formula of binomial coefficients and theorems. There are also multiple properties of binomial coefficients some of which we will need to use in this question to simplify and solve the expression and get its answer. The formulas and properties we will require to solve this question are
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}}\]
Complete step-by-step answer:
Now to start solving this sum you begin with by using the properties of binomial theorem to simplify the LHS of the expression so that once its simplified into a one term expression we can compare both the sides of the equation to get the answer of this question. Hence
\[^{n-1}{{C}_{5}}{{+}^{n-1}}{{C}_{6}}\text{ }<{{\text{ }}^{n}}{{C}_{7}}\]
We can use this by the known property that \[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}}\] here we will take the bigger r and add \[1\] to n to get the simplified coefficient so using this we get
\[^{n}{{C}_{6}}\text{ }<{{\text{ }}^{n}}{{C}_{7}}\]
Now we know the expression of how to write a binomial coefficient so using that we get
\[\dfrac{n!}{6!\left( n-6 \right)!}<\dfrac{n!}{7!\left( n-7 \right)!}\]
The factorial of n of both sides will get cancelled therefore then further simplifying of factorial we get
\[\dfrac{7!}{6!}<\dfrac{(n-6)!}{\left( n-7 \right)!}\]
Now using the factorial of n to simplify we get,(formula for factorial of n is \[n!=n(n-1)(n-2)(n-3)!\])
\[\dfrac{7.6!}{6!}<\dfrac{(n-6)(n-7)!}{\left( n-7 \right)!}\]
Dividing we get
\[7 < n-6\]
Simplifying
\[n>13\]
Now from this expression we can see that the least positive integer this expression is true for is \[14\] and hence the answer for this question is option \[1\].
So, the correct answer is “Option 1”.
Note: Binomial theorem is a way to express the algebraic expansion of the powers of a binomial. It can also be called binomial expansion. Now we must also know that binomial coefficients are the positive integers which usually occur as coefficients in binomial expansion.
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
\[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}}\]
Complete step-by-step answer:
Now to start solving this sum you begin with by using the properties of binomial theorem to simplify the LHS of the expression so that once its simplified into a one term expression we can compare both the sides of the equation to get the answer of this question. Hence
\[^{n-1}{{C}_{5}}{{+}^{n-1}}{{C}_{6}}\text{ }<{{\text{ }}^{n}}{{C}_{7}}\]
We can use this by the known property that \[^{n}{{C}_{r-1}}{{+}^{n}}{{C}_{r}}{{=}^{n+1}}{{C}_{r}}\] here we will take the bigger r and add \[1\] to n to get the simplified coefficient so using this we get
\[^{n}{{C}_{6}}\text{ }<{{\text{ }}^{n}}{{C}_{7}}\]
Now we know the expression of how to write a binomial coefficient so using that we get
\[\dfrac{n!}{6!\left( n-6 \right)!}<\dfrac{n!}{7!\left( n-7 \right)!}\]
The factorial of n of both sides will get cancelled therefore then further simplifying of factorial we get
\[\dfrac{7!}{6!}<\dfrac{(n-6)!}{\left( n-7 \right)!}\]
Now using the factorial of n to simplify we get,(formula for factorial of n is \[n!=n(n-1)(n-2)(n-3)!\])
\[\dfrac{7.6!}{6!}<\dfrac{(n-6)(n-7)!}{\left( n-7 \right)!}\]
Dividing we get
\[7 < n-6\]
Simplifying
\[n>13\]
Now from this expression we can see that the least positive integer this expression is true for is \[14\] and hence the answer for this question is option \[1\].
So, the correct answer is “Option 1”.
Note: Binomial theorem is a way to express the algebraic expansion of the powers of a binomial. It can also be called binomial expansion. Now we must also know that binomial coefficients are the positive integers which usually occur as coefficients in binomial expansion.
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