PE 1 - Permutation & Combination | Modern Math - Permutation & Combination

Answer: Option D

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Explanation :

There are 4 Ss, 4 Is, 2 Ps, 1 M

Case 1: All 4 distinct letters
There is only one way of selecting 4 distinct letters, i.e., S, I, P, M

Case 2: 2 similar letters and 2 other distinct letters
2 similar letters can be selected in 3 ways (i.e., S or I or P) and the remaining 2 distinct letters can be selected from other 3 letters in 3C2 = 3 ways.

∴ Total number of ways = 3 × 3 = 9 ways.

Case 3: 2 similar letters and 2 other similar letters
2 set of similar letters can be selected in 3C2 = 3 ways (i.e., SI or IP or PS)

Case 4: 3 similar letters and 1 other distinct letter
3 similar letters can be selected in 2 ways (i.e., S or I) and the remaining 1 distinct letter can selected from other 3 letters in 3 ways.

∴ Total number of ways = 2 × 3 = 6 ways.

Case 5: All 4 similar letters
4 similar letters can be selected in 2 ways (i.e., S or I)

⇒ Total number of ways = 1 + 9 + 3 + 6 + 2 = 21 ways.

Hence, option (d).

Answer: 176

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Explanation :

In the previous question we calculated number of ways of selecting 4 letters. Now for each of the cases discussed in the previous question we need to calculate the number of ways of arranging the letters.

Case 1: All 4 distinct letters
Number of ways of selection = 1
Number of ways of arrangementss = 1 × 4! = 24

Case 2: 2 similar letters and 2 other distinct letters
Number of ways of selection = 9
Number of ways of arrangements = 9 × $\frac{4!}{2!}$ = 108

Case 3: 2 similar letters and 2 other similar letters
Number of ways of selection = 3
Number of ways of arrangements = 3 × $\frac{4!}{2!2!}$ = 18

Case 4: 3 similar letters and 1 other distinct letter
Number of ways of selection = 6
Number of ways of arrangements = 6 × $\frac{4!}{3!}$ = 24

Case 5: All 4 similar
Number of ways of selection = 2
Number of ways of arrangements = 2 × $\frac{4!}{4!}$ = 2

⇒ Total number of arrangements = 24 + 108 + 18 + 24 + 2 = 176 ways.

Hence, 176.

Answer: Option C

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Explanation :

A number in each of the 5 rings can be selected in 10 different ways.

∴ Total number of combination possible = 10 × 10 × 10 × 10 × 10 = 10⁵

Here, we don’t have to open the lock, we just have to figure out the right combination of lock.

If we consider that the last combination that we get is the right combination, then the maximum combinations we need to try would be  10⁵ – 1. 

∴ Required number of attempts before getting the right combination = 10⁵ – 1

Hence, option (c).

Answer: 126

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Explanation :

We know, n similar objects can be distributed in r different groups in ^n+r-1C_r-1 ways.

∴ 5 similar rings can be worn on 5 (different) fingers of a hand in ^5+5-1C_5-1 = 126 ways.

Hence, 126.

Answer: 1680

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Explanation :

2 of the 6 digits of the number are already known. 

Since all the digits in the number are different, the remaining 4 digits can be any digit except 1 and 0. 

So, these 4 digits can be selected from any of the remaining 8 digits 2, 3, 4, 5, 6, 7, 8 and 9.

∴ Total number of possible telephone numbers = ⁸P₄ = 1680

Hence, 1680.

Answer: Option D

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Explanation :

The number must be greater than 24000 and each digit is to be used exactly once. 

⇒ all the numbers will be 5 digits numbers.

So, numbers starting with 7, 8 and 9 will always be greater than 24000.

In this case the 1^st digit can be selected in 3 ways. The remaining 4 digits can be selected in 4! ways. 

∴ Total number of possible numbers when the 1st digit is 7, 8 or 9 = 3 × 4 × 3 × 2 × 1 = 72.

Now, when the 1^st digit is 2, the second digit cannot be 0. The 2^nd digit can be any of 7, 8 or 9. So, the 2^nd digit can be selected in 3 ways. The remaining 3 digits can be selected in 3! ways. 

So, total number of possible numbers when the 1^st digit is 2 = 1 × 3 × 3 × 2 × 1 = 18

The number cannot start with 0.

∴ Total number of required numbers = 72 + 18 = 90  

Hence, option (d).

Answer: 90

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Explanation :

We will first select the right sock. This can be done in 10 ways.

The left sock can be any except that corresponding to the right leg already selected. So, it can be selected in 9 ways.

∴ Required number of ways = 10 × 9 = 90.

Hence, 90.

Answer: 11

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Explanation :

By selecting 4 vertices we can get exactly one pair of diagonals which will intersect each other. 

∴ The maximum number of points of intersection of the diagonals inside the polygon is same as selecting 4 vertices i.e., ⁿC₄.

⇒ ⁿC₄ = 330

⇒ $\frac{n(n-1)(n-2)(n-3)}{4\times 3\times 2\times 1}$ = 330

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 3 × 4 × 3 × 2

⇒ n(n – 1) (n – 2)(n – 3) = 11 × 10 × 9 × 8

⇒ n = 11

Hence, 11.

Answer: 230

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Explanation :

Connections within the Zone
Number of connections required within 1 zone = ⁵C₂ × 2 = 20
∴ Number of connections required within each of 4 zones = 20 × 4 = 80 connections

Connections across the Zones
Number of connections required between any 2 zones = ⁵C₁ × ⁵C₁ = 25
Number of combination of zones = ⁴C₂ = 6
∴ Number of connections required across the 4 zones = 25 × 6 = 150 connections

⇒ Total number of connections required = 80 + 150 = 230.

Hence, 230.

Answer: 12

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Explanation :

Total number of pairs = ⁿC₂

Number of pairs standing next to each other = n

∴ Number of pairs not standing next to each other = ⁿC₂ – n = $\frac{n(n-3)}{2}$

∴ Total time required = $\frac{n(n-3)}{2}\times \frac{5}{2}$ = 135

⇒ n(n – 3) = 108 = 12 × 9

⇒ n = 12

∴ There were 12 students standing in the circle.

Hence, 12.