A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then b+3 g is equal to ____________.

Solution

Maths sir · 4 steps

Write 168 as a product of ᵇC₃ and ᵍC₂ and test which split actually works.

1. Set up. Choose the boys and the girls independently and multiply. So ᵇC₃ × ᵍC₂ = 168, where b and g are whole numbers.
2. List the values. ᵇC₃ takes only the values 1, 4, 10, 20, 35, 56, 84, … and ᵍC₂ takes only 1, 3, 6, 10, 15, 21, 28, …. Since 168 = 2³ × 3 × 7, check each ᵇC₃ that divides 168.

   b	ᵇC₃	Needed ᵍC₂ = 168 ÷ ᵇC₃	Possible?
   3	1	168	no, 168 is not of the form g(g-1)/2
   4	4	42	no
   8	56	3	yes, g = 3
   9	84	2	no

3. Check. With b = 8 and g = 3: ⁸C₃ × ³C₂ = 56 × 3 = 168. The other values of ᵇC₃, which are 10, 20 and 35, do not divide 168, so this is the only solution.
4. So. b + 3g = 8 + 3 × 3 = 17.

The idea to keep. When a count is given as a product of combinations, factorise the number. Then match the factors against the short list of values each ⁿCᵣ can take.

Answer from the NTA final answer key. Solution by PrepWiser.