Evaluate 72: 72=(7!)/(2! 5!)=21 Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r State the final answer: 21 Answer: 21

Evaluate 103: 103=(10!)/(3! 7!)=120 Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r State the final answer: 120 Answer: 120

Evaluate 92: 92=(9!)/(2! 7!)=36 Recall the binomial theorem: (a+b)^n=_r=0^nnra^n-rb^r State the final answer: 36 Answer: 36

Find the value of 60+66.

Evaluate 60: 60=(6!)/(0! 6!)=1 Evaluate 66: 66=(6!)/(6! 0!)=1 Add the results together: 1 + 1 = 2 State the final answer: 2 Answer: 2

Binomial expansion Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Binomial expansion questions. See exactly how to solve problems on binomial coefficient, nCr, special cases, Pascal's triangle.

binomial coefficientnCrspecial casesPascal's trianglesum of coefficientsexpansion

A-Level70 questionsStep-by-step solutions

Question 1

2 markseasy

Evaluate

\binom{7}{2}

Worked solution

Evaluate \binom{7}{2}
$\binom{7}{2}=\frac{7!}{2!\,5!}=21$
The coefficient \binom{7}{2} is worked out with the nCr formula, which counts the ways of choosing 2 things from 7. You can also read it straight from row 7 of Pascal's triangle.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
State the final answer
$21$
This value is the answer. A quick sanity check: coefficients like this are always whole numbers.

Answer

21

Question 2

2 markseasy

Evaluate

\binom{10}{3}

Worked solution

Evaluate \binom{10}{3}
$\binom{10}{3}=\frac{10!}{3!\,7!}=120$
The coefficient \binom{10}{3} is worked out with the nCr formula, which counts the ways of choosing 3 things from 10. You can also read it straight from row 10 of Pascal's triangle.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
State the final answer
$120$
This value is the answer. A quick sanity check: coefficients like this are always whole numbers.

Answer

120

Question 3

2 markseasy

Evaluate

\binom{9}{2}

Worked solution

Evaluate \binom{9}{2}
$\binom{9}{2}=\frac{9!}{2!\,7!}=36$
The coefficient \binom{9}{2} is worked out with the nCr formula, which counts the ways of choosing 2 things from 9. You can also read it straight from row 9 of Pascal's triangle.
Recall the binomial theorem
$(a+b)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
This single formula generates every term of the expansion. Getting comfortable writing it down first stops us from missing terms later.
State the final answer
$36$
This value is the answer. A quick sanity check: coefficients like this are always whole numbers.

Answer

36

Question 4

2 markseasy

Find the value of

\binom{6}{0}+\binom{6}{6}

Worked solution

Evaluate \binom{6}{0}
$\binom{6}{0}=\frac{6!}{0!\,6!}=1$
The coefficient \binom{6}{0} is worked out with the nCr formula, which counts the ways of choosing 0 things from 6. You can also read it straight from row 6 of Pascal's triangle.
Evaluate \binom{6}{6}
$\binom{6}{6}=\frac{6!}{6!\,0!}=1$
The coefficient \binom{6}{6} is worked out with the nCr formula, which counts the ways of choosing 6 things from 6. You can also read it straight from row 6 of Pascal's triangle.
Add the results together
$1 + 1 = 2$
The question asks for the total, so we simply add the separate coefficients. Keeping them on one line makes the addition clear.
State the final answer
$2$
This value is the answer. A quick sanity check: coefficients like this are always whole numbers.

Answer

2

Question 5

2 markseasy

Find the sum of all the entries in row 5 of Pascal's triangle (the row that begins

1, 5, \dots

Worked solution

Write down row 5 of Pascal's triangle
$1,\ 5,\ 10,\ 10,\ 5,\ 1$
Each row of Pascal's triangle lists the binomial coefficients. Row 5 starts and ends in 1 and is symmetric.
Add the entries in the row
$1 + 5 + 10 + 10 + 5 + 1 = 32$
The question wants the total of the row, so we simply add the numbers. Adding from the outside inwards can make it easier.
Link the total to a power of 2
$2^{5}=32$
The entries in row n of Pascal's triangle always add up to 2^n. This is a useful fact and a good check on our addition.
State the answer
$32$
So the entries in row 5 add to 32. This matches 2 to the power 5.

Answer

32

Unlock 65 more Binomial expansion questions

Create a free account to work through every A-Level Binomial expansion question with instant step-by-step worked solutions, progress tracking and interactive lessons.

Full worked solutions for every question
Interactive lessons and instant feedback
Track your mastery across every topic

Create a Free Account

No card required · Free forever

More Binomial expansion practice

Hard Questions Practice Questions

Binomial expansion Worked Solutions — A-Level Maths

Worked solution

Worked solution

Worked solution

Worked solution

Worked solution

Unlock 65 more Binomial expansion questions

More Binomial expansion practice

Related Pure Maths topics