{"id":666,"date":"2025-05-08T07:27:13","date_gmt":"2025-05-08T07:27:13","guid":{"rendered":"https:\/\/yaveni.com\/blog\/?p=666"},"modified":"2025-05-08T07:27:17","modified_gmt":"2025-05-08T07:27:17","slug":"how-many-3-digit-numbers-are-possible-using-permutations-without-repetition-of-digits-if-digits-are-1-9","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/how-many-3-digit-numbers-are-possible-using-permutations-without-repetition-of-digits-if-digits-are-1-9\/","title":{"rendered":"How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9"},"content":{"rendered":"\n

How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?<\/p>\n\n\n\n

a) 504
b) 729
c) 1000
d) 720<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Correct Answer: d) 720<\/strong><\/p>\n\n\n\n

\n\n\n\n

Explanation:<\/strong><\/h3>\n\n\n\n

We are asked to determine how many 3-digit numbers<\/strong> can be formed without repeating digits<\/strong>, using the digits from 1 to 9<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Understand the Range of Digits<\/strong><\/h3>\n\n\n\n

The digits available are 1 through 9<\/strong>, inclusive. That gives us:<\/p>\n\n\n\n

$$
\\text{Total digits available} = 9 \\text{ digits} \\quad (\\text{Digits: } 1, 2, 3, 4, 5, 6, 7, 8, 9)
$$<\/p>\n\n\n\n

Note: 0 is not included<\/strong>, so we don\u2019t have to worry about a leading zero or its inclusion at all.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Understand the Requirement<\/strong><\/h3>\n\n\n\n

We are forming 3-digit numbers<\/strong> using permutations without repetition<\/strong>. That means:<\/p>\n\n\n\n

    \n
  • The order matters<\/strong> (since 123 is different from 321).<\/li>\n\n\n\n
  • No digit repeats<\/strong> in any number.<\/li>\n<\/ul>\n\n\n\n

    So, we are choosing 3 digits from the 9 available and arranging them in all possible ways.<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Use the Permutation Formula<\/strong><\/h3>\n\n\n\n

    The number of permutations without repetition<\/strong> of r<\/strong> items from n<\/strong> total items is given by:<\/p>\n\n\n\n

    $$
    P(n, r) = \\frac{n!}{(n – r)!}
    $$<\/p>\n\n\n\n

    Here:<\/p>\n\n\n\n

      \n
    • $n = 9$ (digits from 1 to 9),<\/li>\n\n\n\n
    • $r = 3$ (we want 3-digit numbers).<\/li>\n<\/ul>\n\n\n\n

      So,<\/p>\n\n\n\n

      $$
      P(9, 3) = \\frac{9!}{(9 – 3)!} = \\frac{9!}{6!}
      $$<\/p>\n\n\n\n

      Now, calculate:<\/p>\n\n\n\n

      $$
      P(9, 3) = \\frac{9 \\times 8 \\times 7 \\times 6!}{6!} = 9 \\times 8 \\times 7 = 504
      $$<\/p>\n\n\n\n

      Wait \u2014 that gives 504<\/strong>, not 720<\/strong>. What\u2019s wrong?<\/p>\n\n\n\n

      Let\u2019s double-check. We’re being asked how many 3-digit numbers<\/strong> can be formed using digits 1\u20139 without repeating digits<\/strong>.<\/p>\n\n\n\n

      Let\u2019s try to count them step-by-step instead.<\/p>\n\n\n\n

      \n\n\n\n

      Step 4: Count the 3-digit numbers manually<\/strong><\/h3>\n\n\n\n
        \n
      • For the hundreds place<\/strong>, we can use any of the 9 digits<\/strong> (1 through 9) \u2013 since 0 is not in the digit set.<\/li>\n\n\n\n
      • For the tens place<\/strong>, we can use 8 remaining digits<\/strong> (excluding the one already used).<\/li>\n\n\n\n
      • For the units place<\/strong>, we can use 7 remaining digits<\/strong>.<\/li>\n<\/ul>\n\n\n\n

        So:<\/p>\n\n\n\n

        $$
        9 \\times 8 \\times 7 = 504
        $$<\/p>\n\n\n\n

        So the correct answer is actually: a) 504<\/strong>, not d) 720<\/strong>.<\/p>\n\n\n\n

        \n\n\n\n

        \u2705 Corrected Answer: a) 504<\/strong><\/h3>\n\n\n\n

        Final Note:<\/h3>\n\n\n\n

        The answer 720<\/strong> would have been correct if the digits 0\u20139<\/strong> were allowed and repetition was not allowed. But since the digits are only 1\u20139<\/strong>, the number of permutations of 3 digits chosen from 9 (without repetition) is 504<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"

        How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9? a) 504b) 729c) 1000d) 720 The correct answer and explanation is : Correct Answer: d) 720 Explanation: We are asked to determine how many 3-digit numbers can be formed without repeating digits, using the digits from 1 to 9. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-666","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/666","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=666"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/666\/revisions"}],"predecessor-version":[{"id":667,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/666\/revisions\/667"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=666"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=666"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=666"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}