PE 3 - Permutation & Combination | Modern Math - Permutation & Combination

Answer: Option D

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Explanation :

Out of 8 men and 6 women, 9 people have to be selected such that 2 men must attend and there has to be at least 3 men and 2 women.

Since 2 men must attend, now we need to select at least 1 more man and 2 women out of remaining 6 men and 6 women.

Case 1: 1 man + 6 women
Number of ways = ⁶C_{​​​​​​​1} × ⁶C₆ = 6 × 1 = 6 ways

Case 2: 2 men + 5 women
Number of ways = ⁶C₂ × ⁶C₅ = 15 × 6 = 90 ways

Case 3: 3 men + 4 women
Number of ways = ⁶C₃ × ⁶C₄ = 20 × 15 = 300 ways

Case 4: 4 men + 3 women
Number of ways = ⁶C₄ × ⁶C₃ = 15 × 20 = 300 ways

Case 5: 5 men + 2 women
Number of ways = ⁶C₅ × ⁶C₂ = 6 × 15 = 90 ways

⇒ Total number of ways = 6 + 90 + 300 + 300 + 90 = 786.

Hence, option (d).

Answer: Option C

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Explanation :

Total number of 2-digit codes = 10 × 10 = 100

Digits which can be read upside down are 0, 1, 6, 8 and 9.

2-digit code formed using these digits that can be read upside down.

Number of 2-digit codes that can be formed using these 5 digits = 5 × 5 = 25.

Out these 25 codes there are some codes which will actually not create confusion: 00, 11, 88 , 69 and 96

Hence, (25 – 5 =) 20 codes can create confusion.

∴ Codes which will not create confusion = 100 – 20 = 80.

Hence, option (c).

Answer: 45

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Explanation :

Minimum value of x = 3, y = 4 and z = 5.

Let x = a + 3, y = b + 4 and z = c + 5 [a, b, c ≥ 0]

⇒ (a + 3) + (b + 4) + (c + 5) = 20

⇒ a + b + c = 8

∴ Number of integral solutions for x + y + z = 20 is same as number of integral solutions for a + b + c = 8.

⇒ Number of integral solutions for a + b + c = 8 = ^8+3-1C_3-1 = ¹⁰C₂ = 45

Hence, 45.

Answer: 3696

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Explanation :

‘0’ occurs exactly 3 times and 5 occurs exactly once while the number is an even number.

Case 1: First 3 digits are 305
‘0’ has to occur 2 more times and one ‘5’ has already been used.

Case 1(a): Last digit is ‘0’
Now, we have to use one more ‘0’ and other 3 digits can be any of 1, 2, 3, 4, 6, 7, 8, 9

3 0 5     __ __ __ _0_  

One ‘0’ can be placed in 4C1 = 4 ways and remaining 3 digits can be filled in 8 × 8 × 8 = 512 ways.

∴ Total number of ways = 4 × 512 = 2048.

Case 1(b): Last digit is not ‘0’
Now, we have to use two more ‘0s’ to be placed.

3 0 5     __ __ __ _2/4/6/8_  

One ‘0’ can be placed in 4C2 = 6 ways,
Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and 
Remaining 2 digits can be filled in 8 × 8 = 64 ways.

∴ Total number of ways = 6 × 4 × 64 = 1536.

Case 2: First 3 digit are ‘936’

Case 2(a): Last digit is ‘0’
Now, we have to use two more ‘0s’ and one ‘5’. 

9 3 6     __ __ __ _0_  

Two ‘0’ and one 5 can be placed in 4C3 × 3 = 12 ways and remaining 1 digit can be filled in 8 ways.

∴ Total number of ways = 12 × 8 = 96.

Case 2(b): Last digit is not ‘0’
Now, we have to use three more ‘0s’ and one 5.

9 3 6     __ __ __ _2/4/6/8_  

Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and 
Three ‘0s’ and a 5 can be placed in 4 ways,

∴ Total number of ways = 4 × 4 = 16.

⇒ Total number of combinations = 2048 + 1536 + 96 + 16 = 3696.

To make sure that Shubham reaches his friend, he will have to try a maximum of 3696 times.

Hence, 3696.

Answer: Option A

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Explanation :

Let us first calculate the number of ways of arranging 6 cars linearly such that the three lawyers are never together.

3 cars of non-lawyers can be arranged in 3! = 6 ways.

| C | C | C |

Now, the cars of lawyers can be placed in between these 3 cars in ⁴C₃ × 3! = 24 ways

∴ Number of ways of arranging cars = 6 × 24 = 144 ways.

Parking slots ⇒ 1 2 3 4 5 6 7 8 9 10

Number of ways of selecting 6 adjacent parking slots = 5

∴ Total number of ways of parking these 6 cards in 6 adjacent slots = 5 × 144 = 720 ways.

Hence, option (a).

Answer: 36

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Explanation :

For a number to be divisible by 3, the sum of the digits should be divisible by 3.

∴ We need to select 4 numbers whose sum is divisible by 3.

We know 0 + 1 + 2 + 3 + 4 = 10, 

Hence, excluding 1 or 4 will give the sum 9 or 6 (divisible by 3)

Case 1: 4 digits are 0, 2, 3 and 4
Number of 4-digit numbers = 3 × 3! = 18.

Case 2: 4 digits are 0, 1, 2 and 3
Number of 4-digit numbers = 3 × 3! = 18.

∴ Total number of numbers = 18 + 18 = 36.

Hence, 36.

Answer: Option C

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Explanation :

Letters when arranged in alphabetical order – B, E, L, O, W

To reach till the word ELBOW, we will consider all the words coming before it in alphabetical order.

Number of words starting with B = 4! = 24

Number of words starting with EB = 3! = 6

The next word will be E L B O W

∴ ELBOW will be the (24 + 6 + 1 =) 31st word.

Hence, option (c).

Answer: 66666600

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Explanation :

Sum = 11111 × (1 + 3 + 5 +7 + 9) × 4! = 66666600.

Hence, 66666600.

Answer: Option D

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Explanation :

Exactly one common side:
We can chose one common side in 10 ways (any one of the 10 sides of the polygon)

If we chose two adjacent vertices of the polygon, we will get one side which will be common to polygon and the triangle. Now the next vertex should be chose such that we don’t get anymore common sides. Hence, the 3^rd vertex should not be adjacent to the ones already chose.

∴ The 3^rd vertex can be chose in 6 ways.

∴ Total number of such triangles = 6 × 10 = 60

Exactly one common side:
For two sides to be common we need to choose 3 adjacent vertices of the polygon. This can be done in 10 ways. (i.e., ABC, BCD, CDE, …, JAB)

∴ Total number of such triangles = 10

⇒ Total number of triangles = 60 + 10 = 70.

Hence, option (d).

Answer: 136

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Explanation :

Let the number of red, blue and green balls selected be ‘r’, ‘b’ and ‘g’ respectively.

∴ r + b + g = 15

Since the balls of each color are identical, the question is same as distributing 15 identical objects in 3 distinct groups.

⇒ Number of ways = ^15+3-1C_3-1 = 136.

Hence, 136.