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CAT 2025 Lesson : Permutations & Combinations - Binomial Expansion

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6. Combinations or Selections

nCr=n!r!(nr)!=nPrr!^{n}C_{r} = \dfrac{n!}{r!(n - r)!} = \dfrac{^{n}P_{r}}{r!}

nCr=nCnr^{n}C_{r} = ^{n}C_{n - r}

The greatest value of
nCr^{n}C_{r} is r=n2r = \dfrac{n}{2} if n\bm{n} is even; and r=(n+1)2r = \dfrac{(n + 1)}{2} or (n1)2\dfrac{(n - 1)}{2} if n\bm{n} is odd.

Example 38

In how many ways can 44 students be selected from a group of 88?

Solution

Note that here the arrangement does not matter. As it is only selections, we can apply the combinations formula.

Number of ways of selecting 44 students from 8=8 = 8C4=8!4!×(84)!=8!4!×4!^{8}C_{4} = \dfrac{8!}{4! \times (8 - 4)!} = \dfrac{8!}{4! \times 4!}

=8×7×6×54×3×2×1=2×7×5=70= \dfrac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 2 \times 7 \times 5 = 70

Answer:
7070


6.1 Binomial Expansion

nC0^{n}C_{0}  ++  nC1^{n}C_{1}  ++  nC2^{n}C_{2}  ++  ...  ++  nCn^{n}C_{n}  ==  2n2^{n}

nC0^{n}C_{0} ++ nC2^{n}C_{2} ++ 2 nC4^{n}C_{4} ++ ... == nC1^{n}C_{1} ++ nC3^{n}C_{3} ++ nC5^{n}C_{5} ++ ... == 2n12^{n - 1}

When
nn is odd, nC0^{n}C_{0} ++ nC1^{n}C_{1} ++ nC2^{n}C_{2} ++ ... ++ nC(n1)/2^{n}C_{(n - 1)/2} = = nC(n+1)/2^{n}C_{(n + 1)/2} ++ nC(n+3)/2^{n}C_{(n + 3)/2} ++ ... ++ nCn^{n}C_{n} == 2n12^{n - 1}

Example 39

In how many ways can 22 or more people be selected from a group of 77 people

Solution

We need to add all cases where exactly 22 are selected, exactly 33 are selected and so on till all 77 are selected.

Number of ways in which
22 or more are selected == 7C2^{7}C_{2} ++ 7C3^{7}C_{3} ++ 7C4^{7}C_{4} ++ 7C5^{7}C_{5} ++ 7C6^{7}C_{6} ++ 7C7^{7}C_{7}

=27(7C0= 2^{7} - (^{7}C_{0} ++ 7C1)=128(1+7)=120^{7}C_{1}) = 128 - (1 + 7) = \bm{120}

Answer:
120120


Example 40

Rasheed gets 77 shots to fire at a target. He wins 11 point if he hits the target and loses 11 point if he misses the target. If the number of points Rasheed has at the end of 77 shots is a positive number, then in how many different ways could he have managed this?

Solution

He gets positive points only if his hits are more than misses. Thus, his hits should have been 4,5,64, 5, 6 or 77.

Number of ways in which these papers can be selected
== 7C4^{7}C_{4} ++ 7C5^{7}C_{5} ++ 7C6^{7}C_{6} ++ 7C7^{7}C_{7} == 272=26=64\dfrac{2^{7}}{2} = 2^{6} = 64

Answer:
6464


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