A set of friends consists of 7 men and 4 women. Three of the men are brothers: Ali, Ben and Charlie.
- (a)Find the number of different arrangements of the 7 men in a line in which Ali and Ben do not stand next to each other.[3]
- (b)Find the number of different arrangements of the 7 men and 4 women in a line in which all the men stand together and all the women stand together.[3]
- (c)In how many ways can the 7 men and 4 women be divided into a group of 6, a group of 3 and a group of 2 if there are no restrictions?[2]
- (d)The 7 men and 4 women are divided at random into a group of 6, a group of 3 and a group of 2. Find the probability that Ali, Ben and Charlie are all in the same group.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [M1] 7! - k, k an integer ≥ 1.
- [M1] m - 6! × n: n = 1, 2 where m an integer > 6! n = 1 can be implied.
- [A1] = 3600
- [M1] 5! × p where p an integer > 1.
- [M1] q × 6 × 5, q × ⁶C₂, q × ⁶P₂, q × (6×5)/2, where q an integer > 1.
- [A1] = 3600
- [M1] 7! × r, where r is an integer, r > 1 or 4! × s, where s is an integer, s > 1.
- [M1] 7! × 4! × t, where t is an integer, t ≥ 1, t = 1 can be implied.
- [A1] 241920 CAO.
- [M1] ¹¹C₆ × ⁵C₃ × [²C₂], u, v, w = 6, 3, 2 u ≠ v ≠ w.
- [A1] = 4620
- [B1] group of 6: ⁸C₃ × ⁵C₃ × (²C₂) = 560. (560 seen, accept un-simplified).
- [B1] group of 3: ⁸C₆(× ²C₂ × ³C₃) = 28. (28 seen accept un-simplified).
- [M1] Probability they are in same group = (their (560+28)) / their (c) or (their (560+28)) / 4620.
- [A1] = 588/4620 = 7/55, 0.127.
- [B1] Group of 6: 6/11 * 5/10 * 4/9 (= 4/33). (0.1212... seen, accept un-simplified).
- [B1] Group of 3: 3/11 * 2/10 * 1/9 (= 1/165). (0.00606(06...) seen, accept un-simplified).
- [M1] Probability they are in same group = their 4/33 + their 1/165.
- [A1] = 7/55, 0.127.
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