IMOClass 11 › Chapter Test

Permutations and Combinations — Chapter Test

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Q1
The number of ways to select a president, secretary and treasurer from 9 members is:
9P3 = 9×8×7 = 504.
Q2
In how many ways can 6 members of a family be seated at a round dining table?
For circular permutations, n items can be arranged in (n - 1)! ways. Here, (6 - 1)! = 5! = 120.
Q3
How many committees of 4 can be formed from 8 students?
8C4 = 70.
Q4
The number of ways to arrange the letters of SUCCESS is:
SUCCESS has 7 letters with S repeated 3 times and C repeated 2 times, so the number of arrangements is 7!/(3!·2!) = 5040/12 = 420.
Q5
The number of ways to select 3 fruits from 7 different fruits is:
7C3 = 35.
Q6
In how many ways can 5 people be seated on 5 chairs in a row?
The number of ways to arrange n distinct items in n places is n!. Here, 5! = 120.
Q7
Evaluate: 0! + 1! + 2! + 3!
0! = 1, 1! = 1, 2! = 2, 3! = 6. Sum = 1 + 1 + 2 + 6 = 10.
Q8
Three boys and three girls are to be seated in a row. In how many ways can they sit so that no two girls are together?
First, seat the 3 boys: 3! = 6 ways. They create 4 spaces (*B_B_B*). Seat the 3 girls in these 4 spaces: 4P3 = 24 ways. Total = 6 × 24 = 144 ways.
Q9
The number of permutations of 5 distinct objects taken 3 at a time is:
5P3 = 5!/(2!) = 60.
Q10
How many odd numbers less than 1000 can be formed using the digits 0, 2, 5, 7 when repetition of digits is allowed?
1-digit odd: 5, 7 (2 ways). 2-digit odd: Tens place cannot be 0 (3 ways: 2, 5, 7), units must be odd (2 ways), total = 3 × 2 = 6. 3-digit odd: Hundreds place cannot be 0 (3 ways), tens place (4 ways), units place (2 ways), total = 3 × 4 × 2 = 24. Total = 2 + 6 + 24 = 32.
Q11
The value of 8C1 is:
nC1 = n, so 8C1 = 8.
Q12
The value of 16C1 is:
nC1 = n.
Q13
The value of 20C19 is:
20C19 = 20C1 = 20.
Q14
How many ways can 2 boys and 2 girls be selected from 5 boys and 4 girls?
5C2 × 4C2 = 10 × 6 = 60.
Q15
The value of 14C13 is:
14C13 = 14C1 = 14.
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