Ida had 5 cards with matching envelopes - same in design, different in

I didn't really get
The no of ways in which other four cards can be put into wrong envelopes = D (4) = 9
Can you please explain?

Selecting one right envelope = 5 C 1 = 5 ways
Dearrangements of N things can be found by

N! ( 1/2!-1/3!+1/4!.....)
With alternative Negative signs.

Dearrangements of 4 envelopes = 4!(1- 1/1!+1/2!-1/3!+1/4!) = 9
Total ways of arrangement = 120

Probability of 1 envelope placed in right place = 9*5/120 = 3/8

Posted from my mobile device

Ida had 5 cards with matching envelopes - same in design, different in color.

=> Cards : \(C_1\), \(C_2\),\(C_3\),\(C_4\),\(C_5\).

=> Envelopes: \(E_1\), \(E_2\),\(E_3\),\(E_4\),\(E_5\).(Same design, different color).

The probability that exactly one card got into the matching envelope

=> Probability : \(\frac{Desired results }{ Total number of results.}\)

5 cards can be put into 5! ways = 120.

Exact one card correct:

Example : \(C_1\) goes in \(E_1\) (Correct).

=> No of ways one card can be chosen which go into the correct envelope out of 5 cards is = \(^{5}\mathrm{C_1}\) = 5
= 5.

No of ways \(C_2\) goes into the wrong envelope: \(E_3\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_2\) has been put in \(E_3\)).

No of ways \(C_3\) goes into the wrong envelope: \(E_2\), \(E_4\), and \(E_5\) = 3 ways - (Suppose \(C_3\) has been put in \(E_2\)).

No of ways \(C_4\) goes into the wrong envelope: \(E_5\) = 1 way

No of ways \(C_5\) goes into the wrong envelope: \(E_4\) = 1 way

Total ways : 3 * 3 * 1 * 1 = 9

Desired result: 5 * 9

Total results: 120

Probability: \(\frac{5 * 9 }{ 120}\) = \(\frac{45 }{ 120}\) = \(\frac{3}{8}\)

Answer C

The number of ways 1 card finds the correct envelope is:

Total ways-3 correct -2 correct - 0 correct

Total ways is 5!=120

3 correct: 5!/3!2! = 10 ways the 3 correctly inserted can be selected × 1 way the remaining 2 can be inserted incorrectly= 10

2 correct: 5!/2!3! = 10 ways the 2 correct can be selected. The remaining 3 can be seen quickly to have only 2 ways to be inserted incorrectly, eg, A can only be inserted into B or C = 20

0 ways, the difficult one. We know A has 4 incorrect choices and that the total calculation is a multiplication so that the answer is an integer multiple of 4, call it

4X

So the answer must be

(120-1-10-20-4X)/120 = (89-4X)/120

Testing the first answer for example, can 89-4X= 3/16 *120 ? No, because 120 isn't a multiple of 16.

Can it equal 5/8*120 ?

5/8 * 120 = 75

89-75=14 is not a multiple of 4

How about 3/8 ? 3/8 * 120 =45
89-45=44 is a multiple of 4
3/8 is the correct answer

Posted from my mobile device