0 reads0 ratings0 helpful
A (3)-letter code is to be formed from (6) distinct letters. If repetition is not allowed, how many codes are possible?एक (3)-अक्षर का कोड (6) अलग अक्षरों से बनाना है। यदि पुनरावृत्ति नहीं है तो कितने कोड बनेंगे?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (120)(120)
For three positions, choices are (6,5,4), so (6\times5\times4=120). Without repetition, a chosen letter is no longer available.तीन स्थानों के लिए (6,5,4) विकल्प हैं इसलिए (6\times5\times4=120)। बिना पुनरावृत्ति चुना हुआ अक्षर फिर उपलब्ध नहीं रहता।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
