JEE Questions for Maths Permutations And Combinations Quiz 10 - MCQExams.com

Total number of handshakes = ¹⁵C₂.

The required number of ways are 

(10 + 1) (9 + 1) (7 + 1) - 1

= 11 x 10 x 8 - 1 = 879

26 cards can be chosen out of 52 cards, in ⁵²C₂₆ ways.There are two ways in which each card can be dealt,because a card can be either from the first pack or from the second. Hence the total number of ways ⁵²C₂₆ .2²⁶.

It is a fundamental concept.

At least one green ball can be selected out of 5 green balls in 2⁵ - 1 i.e., in 31 ways. Similarly at least one blue ball can be selected from 4 blue balls in 2⁴ - 1 = 15 ways. And at least one red or not red can be select in 2³ = 8 ways.

Hence required number of ways = 31 x 15 x 8 = 3720.

The selection can be made in ⁵C₃ x ²²C₉.[Since 3 vacancies filled from 5 candidates in ⁵C₃ ways and now remaining candidates are 22 and remaining seats are 9]

5 person are to be seated on 8 chairs i.e., ⁸C₅ x 5! or 6720.

[Since 5 chairs can be select in ⁸C₅ and then 5 persons can be arranged in 5! ways]

First omit two particular persons, remaining 8 persons may be 4 in each boat. This can be done in ⁸C₄ ways.The two particular persons may be placed in two ways one in each boat. Therefore total number of ways are = 2 x (⁸C₄).

The total number of two factor products = ²⁰⁰C₂. The number of numbers from 1 to 200 which are not multiples of 5 is 160. Therefore total number of two factor products which are not multiple of 5 is ¹⁶⁰C₂.Hence the required number of factors = ²⁰⁰C₂  - ¹⁶⁰C₂ = 7180.

The number of times he will go to the garden is same as the number of selecting 3 children from 8.Therefore, the required number = ⁸C₃ = 56.

1! + 2! + 3! + 4! = 33

5! = 120, 6! = 720, 7! = 5040

8! = 40320, 9! = 326880.

Thus the two digit of

1! + 2! + ...... + 9! = 1

Also note that n! is divisible by 100 for all n ≥ 10.

∴ term digits of 10! + 11! + .....+49! = 0

∴ term digits of 1! + 2! + ...... + 49! = 1.

The letters appearing without repetitions are H, E, C, I and S, so they can be put on odd places i.e. 5 letters can be placed on 3 places i.e. ⁵P₃ ways 2nd and 4th position can be filled by the letter from M, A and T.

Even places can be filled in two ways.

(i) Choose 1 letter from 3 given letters M, A and T and arrange them in 2! ways

i.e. ³C₂ × 2! ways.

So, the number of ways for even places = ³C₁ + ³C₂ × 2! = 9 ways

Hence he reqd. number of ways = ⁵P₃ × 9 = 540 ways.

Given, total number of points = n and number of collinear points = p. We know that one line has two end points. Therefore total number of  lines = ⁿC₂.Since  p points are collinear, therefore total number of  lines drawn from collinear points = ^PC₂ . We also know that, corresponding to the line of collinearity, one will also be added. Therefore number of  lines = ⁿC₂ – ^PC₂ + 1.