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Permutations and Combinations question

2022 · 28 Jul · Shift 2 · Q36
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Permutations and Combinations question

2022 · 28 Jul · Shift 2 · Q36

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then b+3 g\mathrm{b}+3 \mathrm{~g}b+3 g is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 17

  1. The number of ways to select 333 boys from bbb boys is (b3)\binom{b}{3}(3b​) and the number of ways to select 222 girls from ggg girls is (g2).\binom{g}{2}.(2g​).

So the total number of ways is (b3)(g2)=168.\binom{b}{3}\binom{g}{2}=168.(3b​)(2g​)=168.

  1. Expand the combinations: (b3)=b(b−1)(b−2)6,(g2)=g(g−1)2.\binom{b}{3}=\frac{b(b-1)(b-2)}{6}, \qquad \binom{g}{2}=\frac{g(g-1)}{2}.(3b​)=6b(b−1)(b−2)​,(2g​)=2g(g−1)​. Thus, b(b−1)(b−2)6⋅g(g−1)2=168,\frac{b(b-1)(b-2)}{6}\cdot \frac{g(g-1)}{2}=168,6b(b−1)(b−2)​⋅2g(g−1)​=168, which gives b(b−1)(b−2) g(g−1)=2016.b(b-1)(b-2)\, g(g-1)=2016.b(b−1)(b−2)g(g−1)=2016.

  2. Now use factor values of combinations directly. We need (b3)(g2)=168.\binom{b}{3}\binom{g}{2}=168.(3b​)(2g​)=168. Factorize: 168=56×3=28×6=8×21=7×24.168=56\times 3=28\times 6=8\times 21=7\times 24.168=56×3=28×6=8×21=7×24.

Now check which factors can be written as combinations:

  • (83)=56\binom{8}{3}=56(38​)=56
  • (32)=3\binom{3}{2}=3(23​)=3

Hence one valid solution is b=8,g=3.b=8, \qquad g=3.b=8,g=3. Then (83)(32)=56⋅3=168,\binom{8}{3}\binom{3}{2}=56\cdot 3=168,(38​)(23​)=56⋅3=168, so it satisfies the condition.

  1. Therefore, b+3g=8+3(3)=8+9=17.b+3g=8+3(3)=8+9=17.b+3g=8+3(3)=8+9=17.
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