Chapter 17: Combinations – Exercise 17.1
Combinations – Exercise – 17.1 – Q.1(i)
.png "Combinations – Exercise – 17.1 – Q.1(i)")
Combinations – Exercise – 17.1 – Q.1(ii)
.png "Combinations – Exercise – 17.1 – Q.1(ii)")
Combinations – Exercise – 17.1 – Q.1(iii)
.png "Combinations – Exercise – 17.1 – Q.1(iii)")
Combinations – Exercise – 17.1 – Q.1(iv)
.png "Combinations – Exercise – 17.1 – Q.1(iv)")
Combinations – Exercise – 17.1 – Q.1(v)
.png "Combinations – Exercise – 17.1 – Q.1(v)")
= 5 + 10 +10 + 5 + 1
= 31
Combinations – Exercise – 17.1 – Q.2
Hence n = n
r = 12 and 5
Applying formula
nCp = nCq = n
Then p + q = n
⟹ nC12 = nC5
12 + 5 = n
⟹ n = 17
Combinations – Exercise – 17.1 – Q.3
If nCp = nCq
Then p + q = n
⟹ nC4 = nC6
4 + 6 = n
⟹ n = 10
then 12Cn = 12C10
Applying (i)
.png)
Combinations – Exercise – 17.1 – Q.4
If nCp = nCp
The p + q = n
⟹ nC10 = nC12
10 + 12 = n
⟹ n = 22
Find 23Cn
⟹ 23C22
Combinations – Exercise – 17.1 – Q.5
If nCp = nCr then P + r = n
∴ x + 2x + 3 = 24
3x = 21
x = 7
Combinations – Exercise – 17.1 – Q.6
If nCp = nCq
⟹ p + q = n
also Cx = 18Cx+2
⟹ x + x + 2 = 18
2x + 2 = 18
2x = 18 - 2 = 16
2x = 16
x = 8
Combinations – Exercise – 17.1 – Q.7
If nCp = nCq
The p + q = n
⟹ 15C3r = 15Cr+3
4r + 3 = 15
4r = 15 - 3 = 12
r = 3
Combinations – Exercise – 17.1 – Q.8
8Cr = 7C2 + 7C3
Applying formula
Cancelling 7! from both sides
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Cancelling 8 on both sides
2 × 5 × 3 × 4 × 3 × 2 × 1 = r!(8 - r)!
(3 × 2)(5 × 4 × 3 × 2 × 1) = r!(8 - r)!
⟹ r! = 3!
r = 3
or
r! = 5!
r = 5
Combinations – Exercise – 17.1 – Q.9
80 - 5r = 11r
80 = 16r
r = 80/16
r = 5
r = 5
Combinations – Exercise – 17.1 – Q.10
Cancelling (n - 2)! from numerator and denominator
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