20:[["$","title","0",{"children":"BMSAT Kozhikode 2025 - Let A = \\a, b, c, d, e\\. The no. of subsets that contain exactly 3 of these elements are? | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined The number of ways to se<=ct exactly 3 e<=ments from the set A = \\a, b, c, d, e\\ is found using combinations since order doesn't matter.

The combination formula: C(n,r) = n!r!(n-r)! where n = 5 (total e<=ments) and r = 3 (e<=ments to se<=ct).

C(5,3) = 5!3!(5-3)! = 5!3! 2! = 5 4 3!3! 2 1 = 5 42 1 = 202 = 10

The 10 subsets are: \\a,b,c\\, \\a,b,d\\, \\a,b,e\\, \\a,c,d\\, \\a,c,e\\, \\a,d,e\\, \\b,c,d\\, \\b,c,e\\, \\b,d,e\\, \\c,d,e\\ Therefore, there are 10 subsets that contain exactly 3 e<=ments."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"BMSAT Kozhikode,QA,Modern Math,Permutation & Combination"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"name":"mobile-web-app-capable","content":"yes"}],["$","meta","8",{"name":"apple-mobile-web-app-title","content":"AfterBoards"}],["$","meta","9",{"name":"apple-mobile-web-app-status-bar-style","content":"default"}],["$","meta","10",{"property":"og:title","content":"BMSAT Kozhikode 2025 - Let A = \\a, b, c, d, e\\. The no. of subsets that contain exactly 3 of these elements are? | PYQs + Solutions"}],["$","meta","11",{"property":"og:description","content":"undefined The number of ways to se<=ct exactly 3 e<=ments from the set A = \\a, b, c, d, e\\ is found using combinations since order doesn't matter.

The combination formula: C(n,r) = n!r!(n-r)! where n = 5 (total e<=ments) and r = 3 (e<=ments to se<=ct).

C(5,3) = 5!3!(5-3)! = 5!3! 2! = 5 4 3!3! 2 1 = 5 42 1 = 202 = 10

The 10 subsets are: \\a,b,c\\, \\a,b,d\\, \\a,b,e\\, \\a,c,d\\, \\a,c,e\\, \\a,d,e\\, \\b,c,d\\, \\b,c,e\\, \\b,d,e\\, \\c,d,e\\ Therefore, there are 10 subsets that contain exactly 3 e<=ments."}],["$","meta","12",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/42"}],["$","meta","13",{"property":"og:site_name","content":"AfterBoards | BMSAT Kozhikode 2025 Preparation"}],["$","meta","14",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","15",{"property":"og:type","content":"website"}],["$","meta","16",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","17",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","18",{"name":"twitter:title","content":"BMSAT Kozhikode 2025 - Let A = \\a, b, c, d, e\\. The no. of subsets that contain exactly 3 of these elements are? | PYQs + Solutions"}],["$","meta","19",{"name":"twitter:description","content":"undefined The number of ways to se<=ct exactly 3 e<=ments from the set A = \\a, b, c, d, e\\ is found using combinations since order doesn't matter.

The combination formula: C(n,r) = n!r!(n-r)! where n = 5 (total e<=ments) and r = 3 (e<=ments to se<=ct).

C(5,3) = 5!3!(5-3)! = 5!3! 2! = 5 4 3!3! 2 1 = 5 42 1 = 202 = 10

The 10 subsets are: \\a,b,c\\, \\a,b,d\\, \\a,b,e\\, \\a,c,d\\, \\a,c,e\\, \\a,d,e\\, \\b,c,d\\, \\b,c,e\\, \\b,d,e\\, \\c,d,e\\ Therefore, there are 10 subsets that contain exactly 3 e<=ments."}],["$","meta","20",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","link","21",{"rel":"shortcut icon","href":"https://www.afterboards.in/favicon.ico","type":"image/x-icon","sizes":"48x48"}],"$L2e","$L2f","$L30","$L31","$L32","$L33","$L34","$L35"]

Let $A = \{a, b, c, d, e\}$. The no. of subsets that contain exactly 3 of these elements are?

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