Test Level 1: Permutation & Combination - 2 - CAT MCQ

Number of squares in a board having m rows and n columns (m ≥ n) = m × n + (m - 1) × (n - 1) + (m - 2) × (n - 2) ... (m - n + 1) × 1
Here, m = 10 and n = 9
Number of squares = 10 × 9 + 9 × 8 + 8 × 7 + 7 × 6 + 6 × 5 + 5 × 4 + 4 × 3 + 3 × 2 + 2 × 1
= 90 + 72 + 56 + 42 + 30 + 20 + 12 + 6 + 2 = 330

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 6 x 5! = shall always have 0 in the unit's place.
And the same is the case for 7!, 8! and the succeeding factorials.
Hence, unit's digit shall be the unit's digit in 1 + 2 + 6 + 24 + 0 = 3.

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

We require that the particular person is found in every group of 6 people out of a total of 10. As this person is common to all groups of 6, we will have to consider the number of ways of choosing the remaining 5 persons out of the other 9. This can be done in ⁹C₅ ways = 126 ways.

Odd digits: 1, 3, 5, 7, and 9
Even digits: 2, 4, 6, and 8
Number of ways to fill an odd place = 5
Number of ways to fill an even place = 4

Therefore,
Total number of cases = (5)⁴ × (4)³

For every box, we have three different possibilities of colours.
Therefore, total number of possibilities = 3 × 3 × 3 × 3 × 3

¹⁰P₅ × ⁴P₃ = ⁶P₅ × X × 2
10 × 9 × 8 × 7 × 6 × 4 × 3 × 2 = 6 × 5 × 4 × 3 × 2 × X × 2
9 × 8 × 7 = X then X = ⁹P₃

Total number of arrangements when n members are arranged in a row = n!
Total number of arrangements when n members are arranged in a circle = (n - 1)!
Difference = n! - (n - 1)!
Difference = 10! - 9!

Let 7 occur in the units place. So, the units digit can be placed in 1 way. The tens digit can be placed by the rest of 9 numbers. The hundreds digit can be placed by 6 only, i.e. one way.
As the numbers are between 600 and 700, number of ways = 1 × 9 × 1 = 9
Now, let 7 occur in the tens place. The units place can be filled by any of the numbers from 0-9 except 7.
The hundreds place can be filled in one way (by only 6). So, total 9 ways. 7 can't occur in the hundreds place.
∴ Total number of ways in which numbers can be formed = 9 + 9 = 18

For 1st award: We have 9 options to distribute.
And for 2nd award: Again we have 9 options to distribute (Any actor can get any number of award)
Similarly for each others we have 9 options.
Total number of ways when repetition is allowed = 9 × 9 × 9 × 9 × 9 × 9 = 9⁶
So, total number of ways of distributing the awards = 9⁶

Selecting 4 balls out of 10 balls having at least 1 white ball = Selecting 4 balls out of 10 balls - Selecting 4 balls out of 10 balls having no white ball
= ¹⁰C₄ - ⁶C₄   

= 210 - 15 = 195

For every single shade of a T-shirt, one can make 35 distinct pairs with differently shaded shorts. Since there are 15 shades of T-shirts, number of ways = 15 × 35 = 525

The correct option is C 60
Three digits can be formed in
⁵P₃ ways that is
5!/(5-3)! = 60

Each invitation card can be sent in 4 ways. Thus, 4 × 4 × 4 × 4 × 4 × 4 = 4⁶.

We need to assume that the 7 Indians are 1 person, so also for the 6 Dutch and the 5 Pakistanis. These 3 groups of people can be arranged amongst themselves in 3! ways. Also, within themselves the 7 Indians the 6 Dutch and the 5 Pakistanis can be arranged in 7!, 6! And 5! ways respectively. Thus, the answer is 3! × 7! × 6! × 5!.

An engineer can make it through in 2 ways, while a CA can make it through in 3 ways. Required ratio is 2:3. Option (b) is correct.

Use the property ⁿC_r = ⁿC_n-r to see that the two values would be equal at n = 11 since ¹¹C₃ = ¹¹C₈.

Rearrangements do not count the original arrangements. Thus, 5!/2! – 1 = 59 ways of rearranging the letters of PATNA.

¹⁰P₃ would satisfy the value given as ¹⁰P₃ = 10 × 9 × 8 = 720.

3 × 4 × 4 × 4 = 192