Test: Permutation & Combination- 4 - CAT MCQ

To determine the type of the 1000th fruit, let's analyze the pattern of fruit placement:

1. Sequence of Fruits: The fruits are placed in the following repeating order:

   • Apples
   • Mangoes
   • Oranges
   • Custard apples

2. Number of Fruits Placed: The number of fruits increases by one with each placement:

   • 1 fruit, then 2 fruits, then 3 fruits, and so on.

3. Steps and Cumulative Fruits:

   • Step 1: 1 Apple (Total fruits: 1)
   • Step 2: 2 Mangoes (Total fruits: 3)
   • Step 3: 3 Oranges (Total fruits: 6)
   • Step 4: 4 Custard apples (Total fruits: 10)
   • Step 5: 5 Apples (Total fruits: 15)
   • ...
   • The cumulative total of fruits after each step is calculated using the formula Sn=n(n+1)2S_n = \frac{n(n+1)}{2}Sn​=2n(n+1)​, where nnn is the step number.

4. Finding the Step Containing the 1000th Fruit:

   • Calculate cumulative totals until reaching or surpassing 1000:
     • Step 44: Total fruits = 990
     • Step 45: Total fruits = 1035
   • The 1000th fruit falls within Step 45.

5. Determining the Fruit Type at Step 45:

   • The fruit types repeat every 4 steps:
     • If nmod  4=1n \mod 4 = 1nmod4=1: Apples
     • If nmod  4=2n \mod 4 = 2nmod4=2: Mangoes
     • If nmod  4=3n \mod 4 = 3nmod4=3: Oranges
     • If nmod  4=0n \mod 4 = 0nmod4=0: Custard apples
   • For Step 45, 45mod  4=145 \mod 4 = 145mod4=1, so the fruit is Apples.

6. Calculating the Position of the 1000th Fruit in Step 45:

   • Fruits from 991 to 1035 are placed in Step 45.
   • The 1000th fruit is the 10th apple placed in this step.

Fruits are ordered like 1. Apple 2. Mangoes 3 Oranges 4. Custard apple Fruits arc being pul up first one. then two, then three, then four and so oil.

Mangoes will come like 2, 6, 10, 14 etc. 100th mango will come 2 + 6 + 10 + 14 + 18 + 22 + 26 + 2= 100
100th mango will come in its 8th turn.
Before that apple must have got its 8th turn, orange and custard apple must have got their 7th turn.
Total apples displayed till now = 1 + 5 + 9 + ... + 29= 120
Total oranges displayed till now = 3 + 7+11 + ... +27=105
Total custard apples displayed till now = 4 + 8+ 12 + ... +28= 112
So, total fruits displayed till now (other than mangoes) = 337
So, the position of the 100th mango = 437

Given, S=(1, 2, 3, …,10).

Two numbers from S are to be selected, such that the sum of the numbers selected is a double-digit number.

If one of the selected number is 10, then, the other number can be any one of 1, 2, 3, ..., 9. So, the number of ways =9.

If one of the selected number is 9, then, the other number can be any one of 1, 2, 3, ..., 8. So, the number of ways =8.

If one of the selected number is 8, then, the other number can be any one of 2, 3, ..., 7. So, the number of ways =6.

If one of the selected number is 7, then, the other number can be any one of 3, 4, ..., 6. So, the number of ways =4.

If one of the selected number is 6, then, the other number can be any one of 4, 5. So, the number of ways =2.

⇒ Total number of ways =9+8+6+4+2=29

Hence, the correct answer is 29.

Given a + b + c + d+ e = 20 ...(1)

a + b + c = 5 ... (2)

Given, system of equations is equivalent to a + b + c = 5 ...(3)
and d+ e = 15

Number non-negative integral solutions of equation (3)

Each of the questions can be answered in 4 ways. So, 10 questions can be answered in 4¹⁰ ways.

This is nothing but the application of the concept of number of factors (see Number System).

nⁿ - n! = 232. Use options now.

Now take different cases to fulfill the conditions given.

Let there be n men participants. 

Then the number of games that the men play between themselves is 2×nC2​ and the number of games that the men played with the women is 2×(2n).
∴2×nC2​−2×2n=66 (by hypothesis)
or n2−5n−66=0 or n=11
Hence, the number of participants is 11 men + 2 women =13.

Four persons have chosen to sit on one particular side (assume side A) and two persons on the other side (assume side B). So, we are supposed to select four persons for side A from the remaining 10 persons and remaining six persons will be sitting on side B. Number of ways 4 persons can be selected from 10 persons = ¹⁰C₄

_{Number of ways 6 persons can be selected from the remaining 6 persons = ⁶C6}

_{Number of ways 8 persons can be arranged on side A = 8!
Number of ways 8 persons can be arranged on side B = 8!
Total number of ways = ¹⁰C4 x ⁶C6 x 8! x 8!}

L et S 1, S2, . . . , S8 denote the stations w here the train does not stop. The four stations where the train stops should be at any four of the nine places indicated by cross.

The number of ways in which n things are distributed among n persons such that every person gets one thing is n!

The total number of ways of distributing n things among n people where each person can get any number of things is nⁿ

Therefore, the total number of ways in which n things are distributed among n persons such that at least one person gets nothing is nⁿ−n!

∴nⁿ−n!=232

There is no general method of solving this equation. By trial and error

2²−2!=2

3³−3!=21

4⁴−4!=232

∴n=4

First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosen in 24 ways. So total no. of ways of selection is 32*24 = 768