Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Chapter 17: Combinations – Exercise 17.1 Combinations – Exercise – 17.1 – Q.1(i) Combinations – Exercise – 17.1 – Q.1(ii) Combinations – Exercise – 17.1 – Q.1(iii) Combinations – Exercise – 17.1 – Q.1(iv) 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) Combinations – Exercise – 17.1 – Q.4 If nCp = nCp The p + q = n ⟹ nC10 = nC12 10 + 12 = n ⟹ n = 22 Find 23Cn ⟹ 23C22 = 23 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 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
= 5 + 10 +10 + 5 + 1
= 31
Hence n = n
r = 12 and 5
Applying formula
nCp = nCq = n
Then p + q = n
⟹ nC12 = nC5
12 + 5 = n
⟹ n = 17
If nCp = nCq
⟹ nC4 = nC6
4 + 6 = n
⟹ n = 10
then 12Cn = 12C10 Applying (i)
If nCp = nCp
The p + q = n
⟹ nC10 = nC12
10 + 12 = n
⟹ n = 22
Find 23Cn
⟹ 23C22
If nCp = nCr then P + r = n
∴ x + 2x + 3 = 24
3x = 21
x = 7
⟹ p + q = n
also Cx = 18Cx+2
⟹ x + x + 2 = 18
2x + 2 = 18
2x = 18 - 2 = 16
2x = 16
x = 8
⟹ 15C3r = 15Cr+3
4r + 3 = 15
4r = 15 - 3 = 12
r = 3
8Cr = 7C2 + 7C3
Cancelling 7! from both sides
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!
or
r! = 5!
r = 5
80 - 5r = 11r
80 = 16r
r = 80/16
Cancelling (n - 2)! from numerator and denominator
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 17: Combinations – Exercise 17.3...
Chapter 17: Combinations – Exercise 17.2...
