Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

A blood bank needs 4 people to help with a blood drive. 11 people have volunteered.
Find how many different groups of 4 can be formed from the 11 volunteers.
Answer:

A blood bank needs 44 people to help with a blood drive. 1111 people have volunteered.\newlineFind how many different groups of 44 can be formed from the 1111 volunteers.\newlineAnswer:

Full solution

Q. A blood bank needs 44 people to help with a blood drive. 1111 people have volunteered.\newlineFind how many different groups of 44 can be formed from the 1111 volunteers.\newlineAnswer:
  1. Identify Problem Type: Identify the type of problem.\newlineWe need to find the number of combinations of 1111 people taken 44 at a time, since the order in which the volunteers are chosen does not matter.
  2. Use Combination Formula: Use the combination formula.\newlineThe number of combinations of nn items taken rr at a time is given by:\newlineC(n,r)=n!r!×(nr)!C(n, r) = \frac{n!}{r! \times (n - r)!}\newlinewhere !! denotes factorial, which is the product of all positive integers up to that number.
  3. Plug in Values: Plug in the values for nn and rr.\newlineIn this case, n=11n = 11 (total volunteers) and r=4r = 4 (number of people needed for the blood drive).\newlineC(11,4)=11!4!×(114)!C(11, 4) = \frac{11!}{4! \times (11 - 4)!}
  4. Calculate and Simplify: Calculate the factorials and simplify.\newline11!=11×10×9×8×7×6×5×4×3×2×111! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\newline4!=4×3×2×14! = 4 \times 3 \times 2 \times 1\newline7!=7×6×5×4×3×2×17! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\newlineNow we can cancel out the common terms in the numerator and the denominator.\newlineC(11,4)=11×10×9×84×3×2×1C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}
  5. Perform Calculation: Perform the calculation.\newlineC(11,4)=11×10×9×84×3×2×1C(11, 4) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}\newlineC(11,4)=11×10×9×24×3×2C(11, 4) = \frac{11 \times 10 \times 9 \times 2}{4 \times 3 \times 2}\newlineC(11,4)=11×10×94×3C(11, 4) = \frac{11 \times 10 \times 9}{4 \times 3}\newlineC(11,4)=11×10×34C(11, 4) = \frac{11 \times 10 \times 3}{4}\newlineC(11,4)=11×5×3C(11, 4) = 11 \times 5 \times 3\newlineC(11,4)=165C(11, 4) = 165

More problems from Experimental probability