A container contains 6 red counters, 4 blue counters, and 3 green counters. In how many di… — JAMB UTME

A container contains $6$ red counters, $4$ blue counters, and $3$ green counters. In how many different ways can a selection of $5$ counters be made if the selection must contain at least $2$ red counters, at least $1$ blue counter, and at most $1$ green counter?

Answer: 750 ways

Answer

The total number of different ways to make the selection is 750.

Breaking the problem into disjoint cases based on the number of green counters (0 or 1) and applying the combination formula for red, blue, and green counters in each valid configuration yields

240 + 510 = 750

total ways.

Step-by-Step Solution

Determine the allowable counts for Green (

G

), Red (

R

), and Blue (

B

) counters

Green counters can be 0 or 1. If

G=0

R+B=5

with

R \geq 2, B \geq 1

. If

G=1

R+B=4

with

R \geq 2, B \geq 1

The constraints state

G \leq 1

R \geq 2

, and

B \geq 1

for a total of 5 counters.

Calculate combinations for Case 1 (

G = 0

)

Ways for

(2R, 3B, 0G) = 60

;

(3R, 2B, 0G) = 120

;

(4R, 1B, 0G) = 60

. Total for Case 1 = 240.

Apply the combinations formula

\binom{n}{r} = \frac{n!}{r!(n-r)!}

to each sub-case.

Calculate combinations for Case 2 (

G = 1

)

Ways for

(2R, 2B, 1G) = 270

;

(3R, 1B, 1G) = 240

. Total for Case 2 = 510.

Apply combinations to choose 1 Green counter along with the valid Red and Blue combinations.

Sum all mutually exclusive cases

Total selection ways =

240 + 510 = 750

According to the addition principle of counting, the totals of mutually exclusive cases are added together.

Key Concept

Combinations with Multiple Conditional Constraints

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