CAIE IGCSE Additional Math 11 permutations and combinations

Use this Permutations and Combinations hub to move from factorial and nPr or nCr equations into restricted arrangements and counting problems.

Covered syllabus points

11.1. Distinguish permutations and combinations 11.2. Use n! and nPr/nCr expressions 11.3. Solve arrangement and selection problems

[Maximum number: 4]

The number of permutations of n items taken 4 at a time is equal to $\frac{1}{6} \times$ the number of permutations of 2 n items taken 3 at a time.

Show that n satisfies the equation $3 n^{2}-19 n+20=0$ .

$\frac{n!}{(n-4)!}=\frac{(2 n)!}{6(2 n-3)!}$

B1

Dep on previous B1
B1 for simplification of one side

e.g. 3(n-2)(n-3)=2(2 n-1)
$3 n^{2}-19 n+20=0 \mathbf{A G}$

B1

Dep on all previous B marks

With sufficient detail shown

\begin{array}{r} n(n-1)(n-2)(n-3)= \frac{2 n(2 n-1)(2 n-2)}{6} \end{array}

[Maximum number: 10]

(a)

Find how many different 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8 and 9, if each digit may be used only once in any number.

[ 1 ]

(b)

How many of the numbers found in part (i) are divisible by 5 ?

[ 1 ]

(c)

How many of the numbers found in part (i) are odd and greater than 7000 ?

[ 4 ]

(d)

The number of combinations of n items taken 3 at a time is 92n. Find the value of the constant n.

[ 4 ]

$\frac{n!}{(n-3)!3!}=92n$
B1

n(n-1)(n-2)=552n
M1

$n(n^2-3n-550)=0$ n(n-25)(n+22)=0
M1

n=25
A1 only

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Covered syllabus points

11. Permutations and combinations question 1

11. Permutations and combinations question 2

Question (a)

Question (b)

Question (c)

Question (d)