Permutations with repetitions

How many times can the input of 1.2.2.3.3.3.4 be permutated into four digits, three digits, and two digits without repetition?
Ex:
4 digits = 1223, 2213, 3122, 2313, 4321. . etc
3 digits = 122.212.213.432. . etc
2 digits = 12, 21, 31, 23

I have tried the permutation formula, and it failed.

Correct answer:

n4 =  115
n3 =  43
n2 =  14

Step-by-step explanation:

a1 = 1223; a2 = 1224; a3 = 1232; a4 = 1233; a5 = 1234; a6 = 1242; a7 = 1243; a8 = 1322; a9 = 1323; a10 = 1324; a11 = 1332; a12 = 1333; a13 = 1334; a14 = 1342; a15 = 1343; a16 = 1422; a17 = 1423; a18 = 1432; a19 = 1433; a20 = 2123; a21 = 2124; a22 = 2132; a23 = 2133; a24 = 2134; a25 = 2142; a26 = 2143; a27 = 2213; a28 = 2214; a29 = 2231; a30 = 2233; a31 = 2234; a32 = 2241; a33 = 2243; a34 = 2312; a35 = 2313; a36 = 2314; a37 = 2321; a38 = 2323; a39 = 2324; a40 = 2331; a41 = 2332; a42 = 2333; a43 = 2334; a44 = 2341; a45 = 2342; a46 = 2343; a47 = 2412; a48 = 2413; a49 = 2421; a50 = 2423; a51 = 2431; a52 = 2432; a53 = 2433; a54 = 3122; a55 = 3123; a56 = 3124; a57 = 3132; a58 = 3133; a59 = 3134; a60 = 3142; a61 = 3143; a62 = 3212; a63 = 3213; a64 = 3214; a65 = 3221; a66 = 3223; a67 = 3224; a68 = 3231; a69 = 3232; a70 = 3233; a71 = 3234; a72 = 3241; a73 = 3242; a74 = 3243; a75 = 3312; a76 = 3313; a77 = 3314; a78 = 3321; a79 = 3322; a80 = 3323; a81 = 3324; a82 = 3331; a83 = 3332; a84 = 3333; a85 = 3334; a86 = 3341; a87 = 3342; a88 = 3343; a89 = 3412; a90 = 3413; a91 = 3421; a92 = 3422; a93 = 3423; a94 = 3431; a95 = 3432; a96 = 3433; a97 = 4122; a98 = 4123; a99 = 4132; a100 = 4133; a101 = 4212; a102 = 4213; a103 = 4221; a104 = 4223; a105 = 4231; a106 = 4232; a107 = 4233; a108 = 4312; a109 = 4313; a110 = 4321; a111 = 4322; a112 = 4323; a113 = 4331; a114 = 4332; a115 = 4333;  n4=115
b1 = 122; b2 = 123; b3 = 124; b4 = 132; b5 = 133; b6 = 134; b7 = 142; b8 = 143; b9 = 212; b10 = 213; b11 = 214; b12 = 221; b13 = 223; b14 = 224; b15 = 231; b16 = 232; b17 = 233; b18 = 234; b19 = 241; b20 = 242; b21 = 243; b22 = 312; b23 = 313; b24 = 314; b25 = 321; b26 = 322; b27 = 323; b28 = 324; b29 = 331; b30 = 332; b31 = 333; b32 = 334; b33 = 341; b34 = 342; b35 = 343; b36 = 412; b37 = 413; b38 = 421; b39 = 422; b40 = 423; b41 = 431; b42 = 432; b43 = 433;  n3=43
c1 = 12; c2 = 13; c3 = 14; c4 = 21; c5 = 22; c6 = 23; c7 = 24; c8 = 31; c9 = 32; c10 = 33; c11 = 34; c12 = 41; c13 = 42; c14 = 43;  n2=14



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Showing 2 comments:
Danka
CAN YOU BE MORE SPECIFIC WITH THE ANSWERS?

Math student
Thats wrong:

For anyone wondering why you got the wrong answer, here is the real explanation:

To find the number of permutations without repetition for a given set of numbers, we can use the formula for permutations:

P(n, r) = n! / (n - r)!

Where P(n, r) denotes the number of permutations of n items taken r at a time, and n! represents the factorial of n.

Let's calculate the number of permutations for the input 1.2.2.3.3.3.4.

For four digits (r = 4):
n = 7 (total number of digits)
r = 4 (number of digits to be taken at a time)

P(7, 4) = 7! / (7 - 4)!
= 7! / 3!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 7 * 6 * 5 * 4
= 840

So, there are 840 permutations of four digits without repetition.

For three digits (r = 3):
n = 7 (total number of digits)
r = 3 (number of digits to be taken at a time)

P(7, 3) = 7! / (7 - 3)!
= 7! / 4!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (4 * 3 * 2 * 1)
= 7 * 6 * 5
= 210

So, there are 210 permutations of three digits without repetition.

For two digits (r = 2):
n = 7 (total number of digits)
r = 2 (number of digits to be taken at a time)

P(7, 2) = 7! / (7 - 2)!
= 7! / 5!
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / (5 * 4 * 3 * 2 * 1)
= 7 * 6
= 42

So, there are 42 permutations of two digits without repetition.

9 months ago  1 Like




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