{"id":36562,"date":"2025-06-25T08:05:28","date_gmt":"2025-06-25T08:05:28","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36562"},"modified":"2025-06-25T08:05:30","modified_gmt":"2025-06-25T08:05:30","slug":"compute-p52-enter-an-exact-number-compute-c75-enter-an-exact-number","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/compute-p52-enter-an-exact-number-compute-c75-enter-an-exact-number\/","title":{"rendered":"Compute P(5,2). (Enter an exact number.) Compute C(7,5). (Enter an exact number.)"},"content":{"rendered":"\n

Compute P(5,2). (Enter an exact number.) Compute C(7,5). (Enter an exact number.)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let\u2019s compute each part:<\/p>\n\n\n\n

\n\n\n\n

1. Compute P(5,2)P(5, 2)P(5,2):<\/strong><\/p>\n\n\n\n

This is a permutation<\/strong> problem, which counts how many ways you can arrange 2 items out of 5 in a specific order<\/strong>.<\/p>\n\n\n\n

The formula for permutations is:P(n,r)=n!(n\u2212r)!P(n, r) = \\frac{n!}{(n – r)!}P(n,r)=(n\u2212r)!n!\u200b<\/p>\n\n\n\n

So,P(5,2)=5!(5\u22122)!=5!3!P(5, 2) = \\frac{5!}{(5 – 2)!} = \\frac{5!}{3!}P(5,2)=(5\u22122)!5!\u200b=3!5!\u200b=5\u00d74\u00d73\u00d72\u00d713\u00d72\u00d71=1206=20= \\frac{5 \\times 4 \\times 3 \\times 2 \\times 1}{3 \\times 2 \\times 1} = \\frac{120}{6} = 20=3\u00d72\u00d715\u00d74\u00d73\u00d72\u00d71\u200b=6120\u200b=20<\/p>\n\n\n\n

\u2705 Answer: P(5,2)=20P(5, 2) = 20P(5,2)=20<\/strong><\/p>\n\n\n\n

\n\n\n\n

2. Compute C(7,5)C(7, 5)C(7,5):<\/strong><\/p>\n\n\n\n

This is a combination<\/strong> problem, which counts how many ways you can choose 5 items out of 7 without considering order<\/strong>.<\/p>\n\n\n\n

The formula for combinations is:C(n,r)=n!r!(n\u2212r)!C(n, r) = \\frac{n!}{r!(n – r)!}C(n,r)=r!(n\u2212r)!n!\u200b<\/p>\n\n\n\n

So,C(7,5)=7!5!\u22c5(7\u22125)!=7!5!\u22c52!C(7, 5) = \\frac{7!}{5! \\cdot (7 – 5)!} = \\frac{7!}{5! \\cdot 2!}C(7,5)=5!\u22c5(7\u22125)!7!\u200b=5!\u22c52!7!\u200b=7\u00d76\u00d75!5!\u22c52\u00d71=42\u00d75!5!\u22c52=422=21= \\frac{7 \\times 6 \\times 5!}{5! \\cdot 2 \\times 1} = \\frac{42 \\times 5!}{5! \\cdot 2} = \\frac{42}{2} = 21=5!\u22c52\u00d717\u00d76\u00d75!\u200b=5!\u22c5242\u00d75!\u200b=242\u200b=21<\/p>\n\n\n\n

\u2705 Answer: C(7,5)=21C(7, 5) = 21C(7,5)=21<\/strong><\/p>\n\n\n\n

\n\n\n\n

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

In combinatorics, permutations<\/strong> and combinations<\/strong> are used to count how many ways objects can be arranged or selected.<\/p>\n\n\n\n

A permutation<\/strong> refers to an arrangement of objects where order matters<\/strong>. The formula for the number of permutations of rrr objects from a set of nnn is:P(n,r)=n!(n\u2212r)!P(n, r) = \\frac{n!}{(n – r)!}P(n,r)=(n\u2212r)!n!\u200b<\/p>\n\n\n\n

In our case, P(5,2)P(5, 2)P(5,2) asks how many ways we can choose 2 objects from a group of 5 and arrange them in sequence. We calculate 5!5!5! (which is 120), and divide it by 3!3!3! (which is 6), giving 20 distinct ordered arrangements.<\/p>\n\n\n\n

On the other hand, a combination<\/strong> refers to a selection of objects where order does not matter<\/strong>. The formula is:C(n,r)=n!r!(n\u2212r)!C(n, r) = \\frac{n!}{r!(n – r)!}C(n,r)=r!(n\u2212r)!n!\u200b<\/p>\n\n\n\n

For C(7,5)C(7, 5)C(7,5), we calculate how many ways we can choose 5 items from 7 where order is irrelevant. The value comes out as 21, meaning there are 21 different ways to choose 5 items from 7 without considering their order.<\/p>\n\n\n\n

Understanding the difference between permutations and combinations is important in probability, statistics, and many real-world applications like lottery games, password generation, and arranging teams.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Compute P(5,2). (Enter an exact number.) Compute C(7,5). (Enter an exact number.) The Correct Answer and Explanation is: Let\u2019s compute each part: 1. Compute P(5,2)P(5, 2)P(5,2): This is a permutation problem, which counts how many ways you can arrange 2 items out of 5 in a specific order. The formula for permutations is:P(n,r)=n!(n\u2212r)!P(n, r) = […]<\/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-36562","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36562","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=36562"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36562\/revisions"}],"predecessor-version":[{"id":36578,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36562\/revisions\/36578"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36562"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36562"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}