अंकों 3, 5, 7, 9 पर विचार कीजिए। इन अंकों से 5-अंकीय ऐसी कितनी संख्याएँ बनाई जा सकती हैं, जिनमें इन चारों अंकों में से प्रत्येक अंक हो?

This question was previously asked in
NDA 02/2021: Maths Previous Year paper (Held On 14 Nov 2021)
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  1. 240
  2. 180
  3. 120
  4. 60

Answer (Detailed Solution Below)

Option 1 : 240
Free
NDA 01/2025: English Subject Test
30 Qs. 120 Marks 30 Mins

Detailed Solution

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प्रयुक्त सूत्र:

n! = n × (n - 1) × (n - 2) × ..... × 3 × 2 × 1

5! = 5 × 4 × 3 × 2 × 1 = 120

2! = 2 × 1 = 2

गणना:

चार स्थितियाँ संभव हैं (3, 5, 7, 9, 3), (3, 5, 7, 9, 5), (3, 5, 7, 9, 7), (3, 5, 7, 9, 9)

(3, 5, 7, 9, 3) के लिए

व्यवस्थाओं की संख्या =

⇒ 

⇒ 60

इसी इसी प्रकार सभी 4 स्थितियों के लिए 60 तरीके
⇒ 60 × 4

⇒ 240 

 

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