In the previous section you calculated combinations using the formula $C_{n,k} = \dfrac{n!}{k!(n-k)!}$. That same quantity has its own, more compact notation, which you will use all the time from now on: the binomial coefficient.
Calculate the numerical value of the following combinations applying their defining formula: a) $\binom{9}{3}$ b) $\binom{15}{0}$ c) $\binom{8}{8}$ d) $\binom{12}{10}$
Calculate the numerical value of the following combinations applying their defining formula:
Some values of $\binom{n}{k}$ are worth knowing by heart because they appear constantly.
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Using the symmetry property of combinations, determine the numerical value of $x$ (given that it is different from 5) that satisfies the following equality:…
Using the symmetry property of combinations, determine the numerical value of $x$ (given that it is different from 5) that satisfies the following equality:
You have already seen that each number in Pascal's triangle is the sum of the two above it. Written with binomial coefficients, that observation is a property with a name of its own.
This rule lets you add binomial coefficients without expanding any factorial, as long as they are "neighbours" (same $n$, and consecutive values of $k$).
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Find the exact value of the following sum without needing to calculate each combination individually, applying known properties:…
Find the exact value of the following sum without needing to calculate each combination individually, applying known properties:
Express the sum $\binom{11}{4} + \binom{11}{5}$ as a single combination using Pascal's identity (recurrence relation) and then calculate its numerical value.
Express the sum $\binom{11}{4} + \binom{11}{5}$ as a single combination using Pascal's identity (recurrence relation) and then calculate its numerical value.
Extension: The general term of the binomial theorem
To expand $(a+b)^n$ you do not always need to write out every term: if you are only interested in one (for example, the one containing $x^3$, or the one with no $x$), you can calculate it directly.
To find the constant term (the one with no $x$), set the total exponent of $x$ in that term equal to $0$ and solve for $k$.
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Determine only the numerical coefficient of the term in $x^3$ in the expansion of the binomial $(x + 4)^5$.
Determine only the numerical coefficient of the term in $x^3$ in the expansion of the binomial $(x + 4)^5$.
Calculate and simplify as far as possible the fourth term in the expansion of the binomial $(3x - \frac{y}{2})^7$, ordering the terms according to descending powers of…
Calculate and simplify as far as possible the fourth term in the expansion of the binomial $(3x - \frac{y}{2})^7$, ordering the terms according to descending powers of $x$.
Binomial coefficients can also count paths. If you can only move to the right or upwards on a grid, each shortest path from $A$ to $B$ is a sequence of "right" (R) and "up" (U) moves, and counting paths is the same as counting the ways of arranging the letters R and U in that sequence: a permutation with repetition, which coincides with a binomial coefficient.
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In a Galton board used for probability demonstrations, a ball falls through 6 rows of pegs. At each row, the ball hits a peg and can bounce to the left or to the right.…
In a Galton board used for probability demonstrations, a ball falls through 6 rows of pegs. At each row, the ball hits a peg and can bounce to the left or to the right. Express using a combination the number of different paths that lead to the third bin from the left (considering the leftmost bin as bin zero) and calculate this value.