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The least positive integral value of $x$ which satisfies the inequality ${}^{10}{C_{x - 1}} > 2 \times {}^{10}{C_x}$ is
1. $7$
2. $8$
3. $9$
4. $10$

Answer
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Hint: In this question, we are given the relation of the combination i.e., ${}^{10}{C_{x - 1}} > 2 \times {}^{10}{C_x}$ and we have to find the least positive integral value of $x$. Firstly, apply the combination formula and solve it further until you will reach the point where $x$ will be left on any of the sides.

Formula Used:
Combination formula – A combination is a mathematical technique for determining the number of potential arrangements in a set of objects where the order of the selection is irrelevant. You can choose the components in any order in combinations. Combinations are studied in combinatorics, but they are also applied in other fields such as mathematics and finance.
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Here, $n! = n \times (n - 1) \times (n - 2)...........$

Complete step by step Solution:
Given that,
${}^{10}{C_{x - 1}} > 2 \times {}^{10}{C_x}$
Using the Combination formula,
$\dfrac{{10!}}{{(x - 1)!\left( {10 - (x - 1)} \right)!}} > 2 \times \dfrac{{10!}}{{x!\left( {10 - x} \right)!}}$
$\dfrac{{10!}}{{(x - 1)!\left( {11 - x} \right)!}} > 2 \times \dfrac{{10!}}{{x!\left( {10 - x} \right)!}}$
Also written as,
$\dfrac{1}{{(x - 1)!\left( {11 - x} \right) \times (10 - x)!}} > 2 \times \dfrac{1}{{x \times (x - 1)!\left( {10 - x} \right)!}}$
Cancel the like terms of the sides, we get $\dfrac{1}{{\left( {11 - x} \right)}} > 2 \times \dfrac{1}{x}$
Cross- multiplying both sides,
$x > 2(11 - x)$
$x > 22 - 2x$
Now, add $2x$ on both sides,
$x + 2x > 22 - 2x + 2x$
$3x > 22$
$x > \dfrac{{22}}{3}$
On solving, we get $x > 7.33 \approx 8$
Thus, the value of $x$ which satisfies the inequality ${}^{10}{C_{x - 1}} > 2 \times {}^{10}{C_x}$ is $8$

Hence, the correct option is 2.

Note: In such inequality equation problems one should always remember that in inequality use the same rules as we use to solve normal equality equations. Exception: Change the sign of inequality when multiplying and dividing any negative number into both sides. Also, the combination is a sort of permutation in which the order of the selection is ignored. As a result, the number of permutations is always more than the number of combinations. This is the fundamental distinction between permutation and combination.