Answer: (a)(i) 2160. (a)(ii) 1260. (b) 6399. (c)(ii) \(n=5\).

Answer: (a)(i) 2160. (a)(ii) 1260. (b) 6399. (c)(ii) \(n=5\).

(a)(i) The first digit has 6 choices, since it cannot be 0. After that, the remaining four positions can be filled in

\(6\times5\times4\times3\)

ways. Hence the number of 5-digit numbers is

\(6\times6\times5\times4\times3=2160\).

(a)(ii) For the number to be even, the final digit must be 0, 2, 4 or 6.

If the final digit is 0, the first four digits are chosen and arranged from the six non-zero digits:

\(6\times5\times4\times3=360\).

If the final digit is 2, 4 or 6, there are 3 choices for the final digit. For each such choice, the first digit has 5 choices, and the next three positions can be filled in \(5\times4\times3\) ways. This gives

\(3\times5\times5\times4\times3=900\).

So the total number of even numbers is

\(360+900=1260\).

(b) First count all teams of 7 from 15 people, then subtract the teams with no men.

The total number of teams is \({15\choose7}\), and the number with no men is \({9\choose7}\). Hence the required number is

\({15\choose7}-{9\choose7}=6435-36=6399\).

(c)(i) Use the formulae for combinations:

\({}^nC_3+{}^nC_2=\frac{n(n-1)(n-2)}{6}+\frac{n(n-1)}{2}\).

Put the terms over a common denominator:

\({}^nC_3+{}^nC_2=\frac{n(n-1)(n-2)+3n(n-1)}{6}\).

Factorising gives

\({}^nC_3+{}^nC_2=\frac{n(n-1)((n-2)+3)}{6}\).

So

(c)(ii) Using the result from part (i),

\(\frac{1}{6}(n^3-n)=4n\).

Multiplying by 6 gives

\(n^3-n=24n\).

Hence

\(n^3-25n=0\),

so

\(n(n^2-25)=0\).

Since \(n\ge3\), the only valid solution is

\(n=5\).