Skip to content
Miscellaneous Examples · Example 59

Q.Quality Control: A medical store receives a shipment of 24 infrared temperature guns, including 5 that are defective. Three of these guns are to be sent to a private hospital.

(i) How many selections can be made
(ii) How many of these selections will contain no defective guns?
Sikkim CbseNCERTSubjective· 2mImportance★★★★★est
90% · 113/126 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Quality-control selection: choosing 3 infrared temperature guns from a shipment of 24 (of which 5 are defective, so 19 are non-defective) — total ways to pick 3, and ways to pick 3 with zero defectives.

Since the 3 guns sent are just a group (no ordering/labelling among them), use combinations:

nCr=n!r! (n−r)!^{n}C_r = \dfrac{n!}{r!\,(n-r)!}

where nn = pool size, rr = number selected.

(i) Total number of selections of 3 guns from all 24

  1. n=24n=24 (total guns), r=3r=3 (guns selected).
  2. 24C3=24!3! 21!=24×23×22×21!3!×21!=24×23×223×2×1^{24}C_3 = \dfrac{24!}{3!\,21!} = \dfrac{24\times23\times22\times21!}{3!\times21!} = \dfrac{24\times23\times22}{3\times2\times1}.
  3. Numerator: 24×23=55224\times23 = 552; 552×22=12144552\times22 = 12144.
  4. Divide by 3!=63! = 6: 12144/6=202412144 / 6 = 2024.

(ii) Number of selections with NO defective guns

  1. Non-defective guns in the shipment =24−5=19= 24 - 5 = 19.
  2. To have zero defectives among the 3 chosen, all 3 must come from these 19 non-defective guns: 19C3=19!3! 16!=19×18×173×2×1^{19}C_3 = \dfrac{19!}{3!\,16!} = \dfrac{19\times18\times17}{3\times2\times1}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.