In how many ways can 5 people stand in a line if a particular person must be at the front?
Fix that person; arrange the other 4: 4! = 24
Answer: 24
Permutations and Combinations · Exercise 1 of 5
40 permutations and combinations questions at competitive-exam level: arranging letters with repeats, numbers from digits, circular seating and necklaces, committees, handshakes, diagonals, lines and triangles, and a street map – each solved one step per line with the common trap named.
In how many ways can 5 people stand in a line if a particular person must be at the front?
Fix that person; arrange the other 4: 4! = 24
Answer: 24
How many 4-letter arrangements can be made from the letters of MONDAY without repetition?
6P4 = 6 × 5 × 4 × 3 = 360
Trap: 15 only chooses the letters.
Answer: 360
How many 4-digit PINs can be made from the digits 0 to 9 if digits may repeat?
A PIN may start with 0
10⁴ = 10,000
Trap: 9,000 treats it as a 4-digit number.
Answer: 10,000
How many 4-digit PINs can be made from the digits 0 to 9 if no digit repeats?
10 × 9 × 8 × 7 = 5,040
Trap: 4,536 forbids 0 in front, but a PIN may start with 0.
Answer: 5,040
In how many ways can a team of 11 be picked from 14 players?
14C11 = 14C3 = 364
Trap: 91 is 14C2.
Answer: 364
A team of 11 is picked from 14 players, and the captain must always be included. In how many ways can it be done?
Captain fixed → choose 10 from the other 13
13C10 = 13C3 = 286
Answer: 286
How many 3-letter arrangements can be made from the letters of ENGLISH without repetition?
7P3 = 7 × 6 × 5 = 210
Trap: 343 allows repeats.
Answer: 210
The map shows a grid of streets. A walker goes from A to B by a shortest route, moving only right or up along the streets.
How many shortest routes are there from A to B?
Any route is 3 rights and 2 ups in some order
5!/(3! 2!) = 10
Trap: 6 multiplies 3 × 2.
Answer: 10
The map shows a grid of streets. A walker goes from A to B by a shortest route, moving only right or up along the streets.
How many shortest routes from A to B pass through the junction P?
A to P: 1 right, 1 up → 2 routes
P to B: 2 rights, 1 up → 3 routes
2 × 3 = 6
Trap: 10 counts all routes.
Answer: 6
In how many ways can 4 people sit in a row of 5 chairs?
5P4 = 5 × 4 × 3 × 2 = 120
Answer: 120
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