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Binomial Theorem · Question 18

JEE Main 2026 (Online) 21st January Evening ShiftNumericalMedium📈 Topic in 52% papers+4
1

If ( 115C0 + 115C1 ) ( 115C1 + 115C2 ) ( 115C12 + 115C13 ) = 1314C0 14C1 14C12, then 30 is equal to .

Answer

32

Solution

General factor (r = 0, 1, , 12):

115Cr + 115Cr+1 = 15Cr+1 + 15Cr15Cr 15Cr+1 = 16Cr+115Cr 15Cr+1

16Cr+1 = 16r + 1 15Cr and 15Cr+1 = 15r + 1 14Cr

factor = 16(r + 1) 15Cr+1 = 1615 14Cr

product = ( 1615 )1314C0 14C1 14C12

= 1615

30 = 32

Answer: 32

JEE Main 2026 (Online) 21st January Evening Shift

Keep practising — Binomial Theorem

2

The number of integral terms in the expansion of (5 12 + 7 18 )1016 is:

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  2. 128
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  4. 129
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3

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4

The remainder when ((64)(64) )(64) is divided by 7 is equal to

  1. 4
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5

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6

For an integer n 2, if the arithmetic mean of all coefficients in the binomial expansion of (x + y)2n - 3 is 16, then the distance of the point P (2n - 1, n2 - 4n ) from the line x + y = 8 is

  1. 2
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7

In the expansion of ( [3]2 + 1 [3]3 )n, n N, if the ratio of 15th term from the beginning to the 15th term from the end is 16, then the value of nC3 is

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8

If 12 (15C1 ) + 22 (15C2 ) + 32 (15C3 ) + + 152 (15C15 ) = 2m 3n 5k, where m, n, k N, then m + n + k is equal to :

  1. 20
  2. 19
  3. 18
  4. 21
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9

The sum of all rational terms in the expansion of (2+ 3)8 is :

  1. 16923
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  3. 3763
  4. 33845
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10

If r=19 ( r+32r ) 9Cr = ( 32 )9 - , , N, then ( + )2 is equal to

  1. 27
  2. 81
  3. 18
  4. 9
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11

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