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Chapter 5: Problem 355

The number of arrangements that can be made out of the letter of the word " SUCCESS " so that the all S's do not come together is (a) 60 (b) 120 (c) 360 (d) 420

Short Answer

Expert verified
The number of arrangements that can be made out of the letters of the word "SUCCESS" so that all S's do not come together is 360.

Step by step solution

01

Calculate the total number of arrangements without restriction

The word "SUCCESS" has 7 letters in total, of which there are 3 S's, 2 C's, and 1 U and 1 E. To find the total number of arrangements, we can use the formula: Total arrangements = \( \frac{n!}{p_1! \cdot p_2! \cdots p_k!}\) where n is the total number of letters, and p are the counts of each repeated letter. In this case, we have: Total arrangements = \( \frac{7!}{3! \cdot 2!}\)
02

Calculate the number of arrangements where all S's are together

Now, we will treat all the 3 S's as a single unit, and calculate the number of arrangements, considering the other letters around them. We now have 4 "spaces" to arrange: {SSS}, C, C, U, and E. So, to arrange these, we can use the formula just like before: Arrangements with all S's together = \( \frac{5!}{2!}\)
03

Calculate the difference of Step 1 and Step 2 to find the number of arrangements such that no two S's come together

Now, to find the number of arrangements with no two S's together, we can simply subtract the arrangements found in Step 2 from the total number of arrangements calculated in Step 1. Required arrangements = Total arrangements (Step 1) - Arrangements with all S's together (Step 2) Required arrangements = \( \frac{7!}{3! \cdot 2!} - \frac{5!}{2!}\) Calculating these values, we get: Required arrangements = 420 - 60 = 360 The answer is (c) 360.

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Most popular questions from this chapter

At an election 3 wards of a town are canvassed by 4,5 and 8 men respectively. If there are 20 volunteers then the number of ways they can be allotted to different wards is ? (a) \({ }^{20} \mathrm{P}_{4} \cdot{ }^{20} \mathrm{P}_{5} \cdot{ }^{20} \mathrm{P}_{8}\) (b) \({ }^{20} \mathrm{C}_{4} \cdot{ }^{20} \mathrm{C}_{5} \cdot{ }^{20} \mathrm{C}_{8}\) (c) \({ }^{20} \mathrm{C}_{4} \cdot{ }^{16} \mathrm{C}_{5} \cdot{ }^{11} \mathrm{C}_{8}\) (d) \((1 / 3 !){ }^{20} \mathrm{C}_{4} \cdot{ }^{16} \mathrm{C}_{5} \cdot{ }^{11} \mathrm{C}_{8}\)

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