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:

a_{1} = 1223; a_{2} = 1224; a_{3} = 1232; a_{4} = 1233; a_{5} = 1234; a_{6} = 1242; a_{7} = 1243; a_{8} = 1322; a_{9} = 1323; a_{10} = 1324; a_{11} = 1332; a_{12} = 1333; a_{13} = 1334; a_{14} = 1342; a_{15} = 1343; a_{16} = 1422; a_{17} = 1423; a_{18} = 1432; a_{19} = 1433; a_{20} = 2123; a_{21} = 2124; a_{22} = 2132; a_{23} = 2133; a_{24} = 2134; a_{25} = 2142; a_{26} = 2143; a_{27} = 2213; a_{28} = 2214; a_{29} = 2231; a_{30} = 2233; a_{31} = 2234; a_{32} = 2241; a_{33} = 2243; a_{34} = 2312; a_{35} = 2313; a_{36} = 2314; a_{37} = 2321; a_{38} = 2323; a_{39} = 2324; a_{40} = 2331; a_{41} = 2332; a_{42} = 2333; a_{43} = 2334; a_{44} = 2341; a_{45} = 2342; a_{46} = 2343; a_{47} = 2412; a_{48} = 2413; a_{49} = 2421; a_{50} = 2423; a_{51} = 2431; a_{52} = 2432; a_{53} = 2433; a_{54} = 3122; a_{55} = 3123; a_{56} = 3124; a_{57} = 3132; a_{58} = 3133; a_{59} = 3134; a_{60} = 3142; a_{61} = 3143; a_{62} = 3212; a_{63} = 3213; a_{64} = 3214; a_{65} = 3221; a_{66} = 3223; a_{67} = 3224; a_{68} = 3231; a_{69} = 3232; a_{70} = 3233; a_{71} = 3234; a_{72} = 3241; a_{73} = 3242; a_{74} = 3243; a_{75} = 3312; a_{76} = 3313; a_{77} = 3314; a_{78} = 3321; a_{79} = 3322; a_{80} = 3323; a_{81} = 3324; a_{82} = 3331; a_{83} = 3332; a_{84} = 3333; a_{85} = 3334; a_{86} = 3341; a_{87} = 3342; a_{88} = 3343; a_{89} = 3412; a_{90} = 3413; a_{91} = 3421; a_{92} = 3422; a_{93} = 3423; a_{94} = 3431; a_{95} = 3432; a_{96} = 3433; a_{97} = 4122; a_{98} = 4123; a_{99} = 4132; a_{100} = 4133; a_{101} = 4212; a_{102} = 4213; a_{103} = 4221; a_{104} = 4223; a_{105} = 4231; a_{106} = 4232; a_{107} = 4233; a_{108} = 4312; a_{109} = 4313; a_{110} = 4321; a_{111} = 4322; a_{112} = 4323; a_{113} = 4331; a_{114} = 4332; a_{115} = 4333; n_{4} = 115

b_{1} = 122; b_{2} = 123; b_{3} = 124; b_{4} = 132; b_{5} = 133; b_{6} = 134; b_{7} = 142; b_{8} = 143; b_{9} = 212; b_{10} = 213; b_{11} = 214; b_{12} = 221; b_{13} = 223; b_{14} = 224; b_{15} = 231; b_{16} = 232; b_{17} = 233; b_{18} = 234; b_{19} = 241; b_{20} = 242; b_{21} = 243; b_{22} = 312; b_{23} = 313; b_{24} = 314; b_{25} = 321; b_{26} = 322; b_{27} = 323; b_{28} = 324; b_{29} = 331; b_{30} = 332; b_{31} = 333; b_{32} = 334; b_{33} = 341; b_{34} = 342; b_{35} = 343; b_{36} = 412; b_{37} = 413; b_{38} = 421; b_{39} = 422; b_{40} = 423; b_{41} = 431; b_{42} = 432; b_{43} = 433; n_{3} = 43

c_{1} = 12; c_{2} = 13; c_{3} = 14; c_{4} = 21; c_{5} = 22; c_{6} = 23; c_{7} = 24; c_{8} = 31; c_{9} = 32; c_{10} = 33; c_{11} = 34; c_{12} = 41; c_{13} = 42; c_{14} = 43; n_{2} = 14

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Showing 2 comments:

Danka

CAN YOU BE MORE SPECIFIC WITH THE ANSWERS?

3 years ago Like

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.

1 year ago 1 Like Like