{"id":16596,"date":"2025-06-11T16:40:35","date_gmt":"2025-06-11T16:40:35","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=16596"},"modified":"2025-06-11T16:40:39","modified_gmt":"2025-06-11T16:40:39","slug":"determine-the-value-of-this-15p6-determine-the-value-of-this","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-value-of-this-15p6-determine-the-value-of-this\/","title":{"rendered":"Determine the value of this: 15P6 = Determine the value of this"},"content":{"rendered":"\n

Determine the value of this: 15P6 = Determine the value of this: 15C6 = Determine the value of this: (17\/11) =<\/p>\n\n\n\n

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

Let’s evaluate the following expressions:<\/p>\n\n\n\n

\n\n\n\n

1. 15P615P6 \u2014 Permutation<\/strong><\/h3>\n\n\n\n

Definition<\/strong>:
Permutations represent the number of ways to arrange r<\/strong> items out of n<\/strong> items in order<\/strong>.<\/p>\n\n\n\n

Formula<\/strong>: nPr=n!(n\u2212r)!nPr = \\frac{n!}{(n – r)!}<\/p>\n\n\n\n

Calculation<\/strong>: 15P6=15!(15\u22126)!=15!9!15P6 = \\frac{15!}{(15 – 6)!} = \\frac{15!}{9!}<\/p>\n\n\n\n

Now calculate: 15P6=15\u00d714\u00d713\u00d712\u00d711\u00d710=3,603,60015P6 = 15 \\times 14 \\times 13 \\times 12 \\times 11 \\times 10 = 3,603,600<\/p>\n\n\n\n

\n\n\n\n

2. 15C615C6 \u2014 Combination<\/strong><\/h3>\n\n\n\n

Definition<\/strong>:
Combinations represent the number of ways to choose r<\/strong> items from n<\/strong> items without regard to order<\/strong>.<\/p>\n\n\n\n

Formula<\/strong>: nCr=n!r!(n\u2212r)!nCr = \\frac{n!}{r!(n – r)!}<\/p>\n\n\n\n

Calculation<\/strong>: 15C6=15!6!\u22c5(15\u22126)!=15!6!\u22c59!15C6 = \\frac{15!}{6! \\cdot (15 – 6)!} = \\frac{15!}{6! \\cdot 9!}<\/p>\n\n\n\n

Compute: 15C6=15\u00d714\u00d713\u00d712\u00d711\u00d7106\u00d75\u00d74\u00d73\u00d72\u00d71=3,603,600720=5,00515C6 = \\frac{15 \\times 14 \\times 13 \\times 12 \\times 11 \\times 10}{6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1} = \\frac{3,603,600}{720} = 5,005<\/p>\n\n\n\n

\n\n\n\n

3. 1711\\frac{17}{11} \u2014 Fraction to Decimal<\/strong><\/h3>\n\n\n\n

Calculation<\/strong>: 1711=1.545454545…=1.54\u203e(repeating decimal)\\frac{17}{11} = 1.545454545… = 1.\\overline{54} \\quad (\\text{repeating decimal})<\/p>\n\n\n\n

\n\n\n\n

Explanation (300 Words Approx.)<\/strong><\/h3>\n\n\n\n

In mathematics, permutations and combinations are essential concepts in probability and counting problems. Permutations, denoted as nPrnPr, are used when the order of the selected items matters. For example, arranging 6 people in a line from a group of 15 means each arrangement is different depending on who comes first, second, and so on. The formula for permutations is n!(n\u2212r)!\\frac{n!}{(n-r)!}. For 15P615P6, we use n=15n = 15, r=6r = 6, and find 15!9!\\frac{15!}{9!}, which simplifies to multiplying 15 down to 10, yielding 3,603,600<\/strong> ways to arrange 6 items from 15.<\/p>\n\n\n\n

Combinations, denoted as nCrnCr, are used when order does not<\/strong> matter \u2014 such as choosing 6 lottery numbers from 15, where the order of drawing doesn\u2019t change the outcome. The formula is n!r!(n\u2212r)!\\frac{n!}{r!(n-r)!}, and for 15C615C6, we get 15!6!\u22c59!=5,005\\frac{15!}{6! \\cdot 9!} = 5,005. This tells us there are 5,005 unique ways to choose 6 items from 15 without regard to order.<\/p>\n\n\n\n

Finally, 1711\\frac{17}{11} is a basic fraction. When converted to a decimal, it gives a repeating value: 1.545454…<\/strong>, which is written as 1.54\u203e1.\\overline{54}. Understanding how to convert fractions to decimals is crucial in many areas of math, including statistics and measurement.<\/p>\n\n\n\n

These three topics \u2014 permutations, combinations, and fractional values \u2014 are foundational for solving real-world problems involving probability, arrangements, and number sense.<\/p>\n","protected":false},"excerpt":{"rendered":"

Determine the value of this: 15P6 = Determine the value of this: 15C6 = Determine the value of this: (17\/11) = The correct answer and explanation is: Let’s evaluate the following expressions: 1. 15P615P6 \u2014 Permutation Definition:Permutations represent the number of ways to arrange r items out of n items in order. Formula: nPr=n!(n\u2212r)!nPr = […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-16596","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16596","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=16596"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16596\/revisions"}],"predecessor-version":[{"id":16597,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/16596\/revisions\/16597"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=16596"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=16596"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=16596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}